help-10

[obtuse pebble]

@oak lintel Has your question been resolved?

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[silent fossil]

Anyone that can help me with this boolean matrix? 🙂

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[tender coyote]

Hi I need some help on this volume question, I'm not really sure if I got the right answer.

[obtuse pebble]

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[fierce ginkgo]

Maybe anyone has an idea?

[obtuse pebble]

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@fierce ginkgo Has your question been resolved?

[dark stirrup]

What is M*?

[obtuse pebble]

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[hallow fossil]

Replace the ? inside the Solution of the Following:
line ( 1 , 2 ) , ( 3 , 1 ) = -?/?x +?/?

[supple pumice]

1/2 and 3/2

[hallow fossil]

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[gritty dagger]

x/2 + 8 - 8 = 12 - 8
x/2 = 4
what do i do now?

[supple pumice]

make the x be alone

[earnest elk]

Isolate x

[inland badge]

Remember what I said before

When it's devide you multiply it to move

Opposite of when you multiply

[gritty dagger]

so X/2 = 4/2?

[inland badge]

No

multiply it by 2

[gritty dagger]

or x/2 * 2 / 4 *2

[inland badge]

Yep

[gritty dagger]

ohh

so the answer will be

x = 8

?

[inland badge]

That will isolate x

correct

[gritty dagger]

hmm theyre just so easy but so forgetive

i understand them but i forget them lol

thank u again :D.

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[undone niche]

i feel there is something funky going on with my algebra, the values in the brackets should be lower so i can make it look like the target expression.

please could someone double check?

[obtuse pebble]

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[bleak skiff]

Hello

[bleak skiff]

Sorry i'm still new at this concept

[proper root]

[bleak skiff]

is this function differientiable at 1?

just accprding to picture

[proper root]

You don't include x-axis in your pic

[bleak skiff]

uhm yes

sorry wait

there

[proper root]

ahh, are you familiar with limits?

[bleak skiff]

yes we just started the class about limits

[proper root]

Makes sense with the assignment 😄

You have to check that the slope of the red function = the slope of the blue function at the meeting point, which here is 1

I'm guessing you mean x = 1 in the question

[bleak skiff]

yes

[proper root]

does this make sense?

[bleak skiff]

yes but only according to graph how is it?

or it is not possible?

slope is the same

[proper root]

Can you rephrase this, I'm not quite sure what you mean here

[bleak skiff]

yes

if you look at the graph only, can you say if it is differientiable or not?

[proper root]

Without zooming 100x, it looks like it

[bleak skiff]

it looks like it does? ok

slope on the left and right is -2

so is it?

[proper root]

You need to give me more info

Did you calculate the slope to be -2? and so is what? 😄

[bleak skiff]

i mean slope at x>1 and x<1 is -2

it is the same

so is it differienciable or not?

because we've been told that for a function to be differentiable, the transition need to be pefectly smooth

[proper root]

this is wrong notation. Do you mean limit or? Take a minute to try and formulate a clear question, I'm happy to help

[proper root]

Yes! 😄

[bleak skiff]

this is what happens when you zoomX100

like what you asked

ypu see it's never perfectly smooth

[proper root]

Use limits to find the slope of the red and blue as they x approaches 1

[bleak skiff]

yes i calculated the f(x)-f(1)/x-1 when x-->1

[proper root]

if they're the same, it's differentiable

[bleak skiff]

yes it's both -3

[proper root]

Then it's differentiable in x = 1

[bleak skiff]

so why is the transition not smooth?

[proper root]

it might be

[bleak skiff]

when we zoom?

[proper root]

Guess you haven't zoomed enough, or the program doesn't draw it that precisely

Math doesn't lie ❤️

[bleak skiff]

it's desmos and it said it's not defined when x=1

[proper root]

Ah I see. It's not defined in x = 1

That's because:

The blue function is only valid for x > 1 and the red for x < 1

That's in your definition of the functions on your picture.
Therefore, there is no function value at exactly x = 1

[bleak skiff]

why dont it say it's -3?

[proper root]

My mistake.

You have defined x = 1 as -3 with the green one

[bleak skiff]

yes

[proper root]

Can I see your assignment? Why do you have that green one

[bleak skiff]

yes wait

the original question was asking which values of a, b and c make the function differentiable at 1

i found c=-3

a=-4

b=0

[proper root]

Seems good to me? 😄

Did you just need to double-check the answer or is there another question?

[bleak skiff]

no the question is the one where we need to find a,b,c and I wanted to check on the graph if it was perfectly smooth with these values

this is the derivatives graph(black and blue)

[zenith raft]

i'm not sure that's going to make it differentiable at 1

the derivatives of the top and the bottom piece should agree at 1

[bleak skiff]

"agree"?

[zenith raft]

be equal i mean

[bleak skiff]

ok

so it looks wrong?

[zenith raft]

yea i think so :c

[bleak skiff]

ok let me try again

[zenith raft]

the derivative of the top piece evaluated at 1 is -pi i think

[bleak skiff]

yes i made a mistake

in the formula

[bleak skiff]

so you're telling me -pi = 2 + a

so a should be -pi - 2?

[zenith raft]

ya

and b should be whatever makes 1 + a + b = -3

or something

[bleak skiff]

and then b must be -2 + pi

[zenith raft]

hopefully that's the right number, can't remember lol

[bleak skiff]

yes that's it

[zenith raft]

haha good

[bleak skiff]

so why is the graph still wrong?

[zenith raft]

what's wrong?

[bleak skiff]

not continuous

i put the same values

c=-3

a=-pi-2

b=pi-2

ah no wait

[zenith raft]

this is what mine looks like

[bleak skiff]

yes

ok thanks

srry

[zenith raft]

np ^-^

[bleak skiff]

but why does it always say not defined when x=1?

[zenith raft]

probably just cus you're telling it to plot stuff for x>1 and x<1 or whatever

[bleak skiff]

ah ok it's normal then? why not just say -3?

[proper root]

Good to see it worked out 😄 Seems there were some miscalculations in the first half. Goodnight ^^

[bleak skiff]

good night thaks

[zenith raft]

if it doesn't have any instructions for what to put when x=1, it won't put anything 🤷🏻‍♀️

if i change the > to a >= it will add (1,-3) for the purple plot

but still leave it undefined on the red plot

[bleak skiff]

x=1 you mean but i did put something like that in the definotion

second line in green

[zenith raft]

yea sorry

[bleak skiff]

this

[zenith raft]

yea but

it's plotting everything separately

the purple curve in my picture has nothing defined for it when x = 1 when i just had x>1

and that's all it's saying

it will stay red is undefined when x = 1

it doesn't even know you want all those pieces to be one function

[bleak skiff]

ok if it works like that

thanks

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[timid silo]

How do I do c

I’ve done this

I don’t know how to do if n is not equal to m

[novel knoll]

I would just check the few cases

WLOG let n<=m. If n=0 then only solution is m=0

If n=1 then no solution possible

If n=2 then only m=2 works

If n>2 we see that 2^(n * m)>2^(n+m)

[timid silo]

What’s WLOT

WLOG

[novel knoll]

Without loss of generality

I’m just saying I only have to look at n<=m

Since its symmetric (exact same argument works just swap n and m everywhere if n>m)

[timid silo]

So do I have to show when n is not equal to m

You don’t have to

Because they are symmetric

[novel knoll]

What?

[timid silo]

Because n and m are symmetric

We just have to set them equal to each other right

[novel knoll]

No?

Just ignore the WLOG and consider the two cases n<m and n>m if it confused you

(You will notice you will write the exact same for n<m and n>m)

[timid silo]

I thought I proved that n could b 0 and n could be 2

[novel knoll]

And those are the only solutions as I showed above?

[timid silo]

Wait so what am I missing

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[cloud imp]

.close

[obtuse pebble]

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[ruby field]

hello i need help with this

.close

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[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

@timid silo Use the formula
y2-y1

---

x2-x1

on which coordinates?

A and b

-4--2/2-1

Thats the slope

thx

Whatca get by chance?

-1/2

-2/1, you mean?

Cause thats half

i put it in and it said i was correct

Huh it should be -2/1 or just -2

Werid

thx

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[shut maple]

Hello! Just want to know if my answers are correct. Thanks!!

[cedar mantle]

um 1 and 2 are correct

checking 3

[shut maple]

Thanksssss^^

[cedar mantle]

im assuming 10 rad/s means that the initial angular velocity is 10 rad/s

yep agree with #3 also

[shut maple]

Okay! Thank you so much!!

[cedar mantle]

np!

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[shadow estuary]

ii need help

[violet compass]

with what?

[shadow estuary]

solve for x

all solutions of x

when i do it i get x=3.9599+n2pi

but theres still like more solutions

on the other side of the parabola

[violet compass]

show me how you are doing this

[shadow estuary]

ok

[shadow estuary]

.

@violet compass

@violet compass

@violet compass

@violet compass

@violet compass

@violet compass

@violet compass

@violet compass

@violet compass

@violet compass

!stop

!done

fuck y9u\

.close

[obtuse pebble]

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[gritty carbon]

@drifting wraith hey

[gritty carbon]

you helped me 20 mins ago with a perm and combs problem

any chance you could lend me a hand again

[shadow estuary]

@violet compass whats your problem

you said you would help me

[gritty carbon]

@shadow estuary i have this channel g

[violet compass]

never promised anything, don't hold anything against it x

[shadow estuary]

@violet compass ok maybe i overreacted

sorry

[gritty carbon]

need help with 1 b

[drifting wraith]

it's the same idea, you still have two colors

ace and non ace

[obtuse pebble]

@gritty carbon Has your question been resolved?

[gritty carbon]

erm

so

it would be

4 aces, and then no ace

[drifting wraith]

yes

[gritty carbon]

5C4

for one part?

[drifting wraith]

no

[gritty carbon]

52 C 4

[drifting wraith]

4C4

[gritty carbon]

shit

[drifting wraith]

there's 4 aces in the deck

[gritty carbon]

4C4 * 48 C 48

[drifting wraith]

48C1

[gritty carbon]

oh right

okay

thanks

.close

[obtuse pebble]

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[timid silo]

You leave Point A at noon, drive 140km to Point B, take a 1/2 hr break and return to Point A at 3:15pm. Traffic is busier on the way back so your average speed is 10km slower. What was your average time?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

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

@timid silo Has your question been resolved?

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[small moth]

I'm having some more trouble with combinatorics. Discrete math is not my forte. My current problem is in regard to a line of 12 people. I know that there are 479,001,600 total ways for the people in the line to be lined up. The question asks "How many ways are there for twelve distinct people to stand with Person A (who we'll call Noga) standing directly in front of Person B (who we'll call Peter)? I have no idea how to start this question.

[gritty carbon]

The wording on this question is a little funky

In other words, person A and B are grouped correct?

[small moth]

I'll try to deliver it a bit more clearly... one second...

Twelve people are standing in a line. Two of these people are Noga and Peter. How many possible lines can be made with Noga standing directly in front of Peter?

yes, they are grouped

[gritty carbon]

okay

So usually you'd approach getting all the solutions by simply doing 12!

for 12 individual people

But the way you should think about it now, is not as people but entities

So if the question were to ask

How many ways can twelve people be arranged in a row

That's twelve separate entities, so therefore it's just 12!

But in this case

We have two people grouped

So in total, we actually have 11 entities

[small moth]

so this would be 11!?

[gritty carbon]

Well the question implies that Noga and Peter have to be ordered in a specific way

so it should just be 11!

but

If the question were to simply say

If Noga and Peter were grouped (it doesn't matter who's standing in front of who), you'd have to multiply it all by 2! again to account for the fact that Noga and Peter can move around in their own group

So noga could be behind peter, or peter could be behind noga

But in your question, it states that Noga has to be infront

So it should just be 11!

[small moth]

thank you!

[gritty carbon]

No worries, let me know if it's the right answer 😄

[small moth]

I won't know until tomorrow afternoon 😅

[gritty carbon]

I'm actually studying combinatorics right now for my exam in two days haha

Doing all this revision

[small moth]

Wuff. My exam is next week

[gritty carbon]

Anyway, best of luck with the topic, it's super disgusting lmao

[small moth]

oh I know

this question actually has multiple parts though. I'm not sure if I can solve them the same way

[gritty carbon]

What grade are you doing this for btw?

Send me the other parts

Good revision for me as well haha

[small moth]

Sophomore level mathematics

[gritty carbon]

Ah right on

so around 15-16

years old

[small moth]

The next part asks "in how many such lines do noga and Peter stand next to each other regardless of order?"

[gritty carbon]

Oh

That's what i was just explaining haha

[small moth]

no no, University sophomore

[gritty carbon]

Bruh, I'm not familiar at all with the us education system

[small moth]

Okay, now that you've said that, I get it

[gritty carbon]

So yeah

Now you have to account for the fact that Noga and Peter can move around in their little group of two

[small moth]

For me it was:
Primary School
Grades 1-3
Intermediate School
Grades 4+5
Middle School
Grades 6-8
High School
Grades 9-12
Freshman, Sophomore, Junior, Senior
University
Grades 13-16
Freshman, Sophomore, Junio, Senior

but it varies by state

[gritty carbon]

That's odd lmao

In australia it's a lot more simple

it's just

Preschool which is basically before 5 years old

Then Kindergarten to Year 6 (Primary School)

Then Year 7 - Year 12 (high school)

then uni

In Year 12 atm

[small moth]

We have preschool and kindergarten too. Some folks go to "Elementary School" which is 3-6, and then "Junior High" which is 7-8

[gritty carbon]

yall got too many schools

[small moth]

I'm actually looking into emigrating to Australia with my girlfriend lol

[gritty carbon]

oh nice

what part?

[small moth]

Not sure yet. For now we're just trying to get away from the US's 2 party system. STV makes too much sense haha

[gritty carbon]

Well unless you got decent money avoid sydney

You'll get rinsed haha

Everything here costs so much

[small moth]

We're gonna get rinsed anyways. Emigrating anywhere from the US can cost close to 35k for two people

[gritty carbon]

Ouch

As for residency, how you handling that?

[small moth]

I have to graduate uni first. Then to emigrate and get citizenship I need to become a Nationally (or internationally) recognized expert in my field and considered valuable to the country

[gritty carbon]

Well, it's best you get stuck into the combinatorics then

Any more questions on the paper?

[small moth]

yeah. Next says "In how many such lines does Noga stand somewhere ahead of Peter?"

[gritty carbon]

Oh yeah these ones are cheeky

So

Think about it like this

Noga, can either be infront or behind of Peter

And if you were to add all the possibilities

It's just a 50% chance he's in front

Or 50% change he's behind

So the way to solve this, is to just get the total possibilities and divide it by 2

$\frac{12!}{2}$

[warm shale]

splooze

[small moth]

Man, I feel dumb as

[gritty carbon]

No, like you literally won't think of that unless you've been told it

[small moth]

[gritty carbon]

The first time I got that question I was writing down like 30 possibilities

And then the answer was just like

divide it all by two

so

Remember that and it'll save you a lot of time lmao

[small moth]

HAHA will do

[gritty carbon]

Because in my case, instead of being infront or behind

It was to the left or the right

So same thing applies

Any other questions ?

[small moth]

Last one says "Three people between Noga and Peter" Is this 1/4 * 12!?

that doesn't feel right

[gritty carbon]

ah yeah these one's i don't like

so the way i approach these ones is i draw a little diagram

instead of being in a line, i'm just going to write it like they're in a row

so to the left or right

so thing of it like this

here is our row of people

i can't do multiple _

okay i'll just write a letter for the placements

here is our row

O O O O O O O O O O O O

twelve spots

N O O O P O O O O O O O

my mouse is broken one sec

[small moth]

you're all good lmao

[gritty carbon]

N O O O P O O O O O O O
O N O O O P O O O O O O
O O N O O O P O O O O O
O O O N O O O P O O O O
O O O O N O O O P O O O
O O O O O N O O O P O O
O O O O O O N O O O P O
O O O O O O O N O O O P

Okay so here's a visual of all the separations

Which in total is

8

But we have to times that by two because

If you switch around the N and the P's

That's still a different order, so we have to account for that

So

We would then do

16 * 10!

I believe

[small moth]

hm okay. That makes a lot of sense

[gritty carbon]

Because we have already identified where the N and the P's are

So we don't have to account for that order

that's what the 16 is doing for us

Now we just have to account for all the possibilities where the other 10 people can be standing

which is where the 10! comes into play

Hopefully that makes sense, I'm still not too comfortable with all that stuff haha

[small moth]

I totally get you lol. I've been programming for 3 years now and this kind of stuff is still lost on me

Thank you so much for your help

[gritty carbon]

Ay nice, I've been programming 3 years as well haha

Np, if you have any more questions lmk

[small moth]

Will do :]

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[primal mango]

3. Note that after 5715 years a given amount of carbon-14 will have decayed to half the original amount.

a. Find the exponential decay model for carbon-14. Round the decay constant to 6 decimal places.
To be honest I don't think I was taught this so I'm not sure but from what I've found online it seems like it could be: (not sure how to enter subscript but imagine there is a 0 in subscript after each y)
0.5y=ye^-0.000121t

[obtuse pebble]

@primal mango Has your question been resolved?

[primal mango]

no

<@&286206848099549185>

[fierce lagoon]

,calc 1/5715

[warm shale]

Result:

1.7497812773403e-4

[fierce lagoon]

Gee thanks

I love scientific

$$H(t) = a_0 e^{-dt}$$

$d$ is your decay constant. You wanna solve for half of $a_0$, so:

$$0.5 a_0 = a_0 e^{-d(5715)}$$

[warm shale]

Umbraleviathan

[obtuse pebble]

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[timid silo]

Is this correct? I tried using the format of a similar example where it was a 15x20 rectangle with 26 points and proving there exists at 1 pair of points where the distance is equal to 5

[timid silo]

Here the question is this

[narrow vault]

can you recall the pigeonhole principle

[timid silo]

yes thats what i was trying to use here

if theres k items and k boxes where k items > k boxes then there will be a box with > 2 items

this question is just kinda confusing i went off an example that i described but i honestly dont really get it

[narrow vault]

oh i didn't scroll up

your construction of the 20 equal area partitions is correct

but your proof is a little shoddy

[timid silo]

yea i figured, i wasnt really that convinced by it since drawing the points in all the rectangles i end up with some spaces unfilled? so i was kinda confused on how this works

[narrow vault]

by partition and PHP you can show there exists at least 1 rectangle containing 2 points

so now what's the largest possible distance between the two points in one rectangular partition

[timid silo]

it would be the diagonal distance in the entire rectangle right? so square root of 3^2 and 4^2 so 5

but i gotta show its less than 5

[narrow vault]

ehh actually when you're being asked to prove <5 instead of <= 5 your professor's argument isn't sufficient

[timid silo]

i see

hm

how would i go about proving <5 in that case

[narrow vault]

so showing at least a row/column of 4 rectangular regions in the 5x4 grid of partitioned areas must contain 5 points and the 5 points is forced to be one of 2 configurations (should be easy to force) then showing each configuration leads to an impossible construction will work but it feels janky

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Is there another method that’s more efficient?

[narrow vault]

off the top of my head no, when PHP doesn't immediately work you mostly resort to casework

at least in this case it's clear your prof intended for the condition to be <5 since it's immediately from PHP :p

[timid silo]

So should I edit my proof or is it sufficient enough for the pigeon hole principle ?

[narrow vault]

when you're being asked to show <5 PHP isn't enough by itself

because yes you have 2 points in the same rectangle but you can (barely) place those 2 points 5 inches away from each other so you're not immediately getting a contradiction

[timid silo]

What should I add then

[narrow vault]

what i said above, show instead that there exists a row of 4 rectangle containing 5 points (or a column of 5 rectangles contains 6 points, either way) and what its distribution must be

then by casework show that they lead to impossible constructions to place 21 points

[timid silo]

Oh right

I’m just a bit confused on how it should look like

Sorry for the bad pic but is this what u mean?

I have 6 points there

Just kinda confused on where to go from here

Do I just add random points everywhere

[narrow vault]

they immediately fail the condition of points being distance <5 apart since your rectangles are 3x4

[timid silo]

Yea I was gonna say I just realized 😭🤣

Sorry ab that it’s late lol

Okay yea this makes sense thank u sm!!

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[timid silo]

6.b

[timid silo]

Is ma method goin ryt?

And calculations

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185> pls help

[humble marten]

dude

specify ur problem

we aint gonna check allat

[timid silo]

Ok

[timid silo]

I used the quadratic eqn?

Should I use it? Or go for profit maximisation method by frst order conditon

I mean if i used any of it would it be correct?

@humble marten

[humble marten]

for which problem

[timid silo]

For 6.b

[timid silo]

.

Question

[timid silo]

Check here I used quadratic eqn

Now I'm gettin confused since the no.s gettin complex

[humble marten]

ok?

what r u confused ab

[timid silo]

I mean i forgot how to do LCM as well💀💀

[humble marten]

List the first several multiples of each number.
Look for multiples common to both lists. If there are no common multiples in the lists, write out additional multiples for each number.
Look for the smallest number that is common to both lists.
This number is the LCM.

[timid silo]

Ok

Thanks

[timid silo]

Can u try on ur own pls

Jst for dis

Ma exams in two days

[humble marten]

im about ot go to bed, sorry

[timid silo]

Ok then

[humble marten]

good luck

[timid silo]

Np

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[empty pollen]

how would I do this question?

[narrow vault]

draw a diagram

[empty pollen]

ive been trying but im not sure where to start

[narrow vault]

you can pretty much figure out what to draw sentence by sentence

[empty pollen]

i dont get how if its on top of the building it can be 120m horizontally from the base of the building

[narrow vault]

that's not what the sentence says

[empty pollen]

ohhh

it would be tan37 (x/120) - tan35(x/120) right

[narrow vault]

using different variables would help

[empty pollen]

120 x (tan37 - tan35) would work right

thanks for the help

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[peak hamlet]

Is my working out at question b.) correct?

[royal basin]

,w d/dx (x^3 - (9/2)x^2 + 6x + 5)

[warm shale]

Results provided by WolframAlpha

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

Requested by Ann#0413

[royal basin]

ok yeah you're good

[obtuse pebble]

@peak hamlet Has your question been resolved?

[peak hamlet]

Alright thanks

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[formal solstice]

Hi, how can I solve this Laplace transform?

[timid silo]

Write down the definition @formal solstice

[formal solstice]

So, f(t) = (5e^2t - 3)^2 here?

[timid silo]

yep

[formal solstice]

Oh ok. Thanks.

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[royal basin]

what is troubling you?

[a] I have no idea where to begin.
[b] I have the beginnings of an attempt, but got stuck midway not knowing what to do.
[c] I have gotten to an answer, but it doesn't match the answer key / is getting rejected / I am told it is wrong.

which of these best describes you

not answering my question here.

yes, of course the goal is to solve for x.

i want to know where your confusion lies.

[royal basin]

.

[timid silo]

Very unique way of helping, pretty rewarding

[royal basin]

okay

right

in this case

may i suggest as a first step that you multiply both sides by (x-1)(x-2)?

do it and show what you get.

if you feel that you can continue on your own afterward, say so.

[timid silo]

Oh

Big oopsie

[royal basin]

[tranquil quiver]

so you are cheating

[royal basin]

no

[tranquil quiver]

yes

[royal basin]

you said that you are in a test

[timid silo]

I mean hey, you are honest at least gotta give it to you

[royal basin]

we do not help cheat on tests

[tranquil quiver]

in what world are you allowed to get the answers from other people

next time get help for the revision work

this is comedic

[timid silo]

Just take the L dude

[tranquil quiver]

no one is going to help you

okay you do you

[obtuse musk]

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[steep trellis]

i am confused

[steep trellis]

tan 30 = tan theta ?????

[static beacon]

$tan(3\theta)$

[warm shale]

Mortta

[static beacon]

not tan30

[steep trellis]

oh oops

[static beacon]

how r u making that mistake lol

[steep trellis]

oh wait

[timid silo]

You can rewrite it as tan(2x+x) and apply trigonometric addition identities. I rewrote theta as x for simplicity

[steep trellis]

oh alr mb mbm xddd

thanks

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[hot wing]

integral of y=x from x = 0 to x = 1

[hot wing]

what does this look like in equation?

latex i mean

[wary kettle]

#latex-help

[warm shale]

ΣAC

[static beacon]

integrate it

[obtuse pebble]

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[exotic ledge]

can the existence of 0 theoreme be applied on the second interval

can me say that f(x) is continous on [0,+inf)

at 0 sure

but i'm doubltful at inf

pls tag me

[zenith raft]

solution in I_2 seems easier, x^4 -2x is 0 when x=0

and it will go off to infinity as x increases

so it will have to cross 1

[exotic ledge]

theoreticaly it has a 0

bcs f(x) = -1

so it has to cross 0 somewhere

this is not difficult to conceptualize

but can we apply this theoreme?

bcs f(x) contious is a must to apply this

[zenith raft]

polynomial functions are continuous

[exotic ledge]

ok

so whb y=x^2 can we apply weirstrass theoreme?

[zenith raft]

what’s that?

[exotic ledge]

on I=[o,+inf)

it means that f(x) achieves a max and min on this interval

if continous in this interval

[zenith raft]

oh

[exotic ledge]

this is a polynomial so it must be conitnous righ?

[zenith raft]

that’s for compact intervals

[exotic ledge]

my book says that you can't apply this theoreme on this interval

bcs it's not bounded

[zenith raft]

yes

[exotic ledge]

no the problem is inf

[zenith raft]

not compact

[exotic ledge]

but how is this different from the first exercise

where it was continous

[zenith raft]

that’s just intermediate value theorem

[exotic ledge]

the condition is the same f(x) has to be continous

[zenith raft]

not weierstrass theorem

[exotic ledge]

but they have the same conidition

[zenith raft]

you can just choose something in (0,infty) for ivt

like 10

anything that makes x^4 - 2x >= 1

idk what you mean by “same condition”

[exotic ledge]

yeah yeah your right

in the second case the function doesn't get a max

[exotic ledge]

by "the function has to be continous" to apply both theoremes

[zenith raft]

ok but they are pretty different theorems lol

[exotic ledge]

i was just concerned by the condition

if i have to prove things in an exam

[zenith raft]

you have a lot of choice in how you apply them to get information out of a function

just having a non-compact interval in the question is OK

they can still be helpful

[exotic ledge]

ok then

thank you

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[zenith raft]

np ^-^

[exotic ledge]

.reopne

.reopen

hey

[obtuse pebble]

✅

[exotic ledge]

i got so confused

is a^2 +b^2 the same as (a+b)^2

[sage geode]

No

[stable rain]

nein

consider a,b=1

a^2 +b^2 = 1+1=2

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

[exotic ledge]

but the formula for the expansion is for a^3 +b^3 right

i was talking about 3rd power sorry

[stable rain]

um

its a bit diff

its

a^3+b^3=(a+b)(a^2-ab+b^2)

[exotic ledge]

yeah that's what i was talking about

but is there a formula for (a+b)^3?

[tidal hawk]

a^3 + b^3 + 3ab(a+b)

or just expand as per normal

[obtuse pebble]

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[mossy jolt]

Combined Variation

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[torpid notch]

how can i discover MU?

[torpid notch]

i dont know how to get the difference between MU

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[wintry stream]

can someone please explain why this is wrong

[obtuse pebble]

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[timid silo]

if a standing wave is formed from two waves interfering. how would one find the node and anti-node points using the wave equation?

[timid silo]

if the node and anti node are points on the wave that stand still, could you let t = 0 and do something?

[obtuse pebble]

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[solemn ravine]

you know how a metric is defined pointwise? As in, it is the distance between two points? Is there anything wrong with plugging in sets instead? Like, if $d : \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$ is the euclidean metric, $x \in \mathbb{R}^2$ and $S \subset \mathbb{R}^2,$ then is the expression $d(x,S)$ valid?

[warm shale]

Joy!

[solemn ravine]

i am assuming that yes it is fine since d is the shortest distance between the two. Hence d(x,S) is well-defined, but there could be something im missing?

[royal basin]

d(x,S) is the infimum of d(x,s) for s in S, but yes.

greatest lower bound

[solemn ravine]

ah thanks, i was looking for a good way to write down d(x,S)

least upper bound = supremum

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[hybrid veldt]

Wouldnt the area be 187? 17x22/2

[fading quest]

Its height times base not height times hypotenuse

[hybrid veldt]

ah okay lol, had h confused with hypotenuse xD

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[outer python]

77

Yeah

K

How

K

Ok

It's correct

Tbh I don't

K

No

Yeah I get it

I got it

Ncr

Yeah

Yeah

Yeah

Yeah

2³

Yeah

Yes

Yeah

Yes

Thank u very much

K

[obtuse pebble]

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[vocal matrix]

im trying to find the side c using law of sines

but when I type it in to my calculator I keep getting different answers from wolfram

8sin(70)/sin(47) = 10.27 correct?

I have a ti89 titanium

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[dense reef]

Hello, does anyone know how to solve this one? I could do a similar question but my mind melts when I have to rationalize

[novel knoll]

Multiply and divide by the conjugate

[dense reef]

I know how to rationalize, I just cant figure out how to apply it to this question

[novel knoll]

Why not

[dense reef]

I can't get it to equal the form in the denominator

sqrt(bx+Ch) + (sqrt(x))

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[gritty harness]

Is the original question

[gritty harness]

My working so far

Please can someone help

[gritty harness]

Is the original question

Really ?

If you can explain it ,it would be much appreciated @timid silo 😁

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@gritty harness Has your question been resolved?

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[obsidian egret]

hi ik this isnt too complicated but im just really tired and its been a while since ive done anything like this please could i get some help?

[tardy epoch]

https://www.cuemath.com/measurement/volume-of-rectangular-prism/

https://www.cuemath.com/measurement/volume-of-cylinder/

[obsidian egret]

thank you

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[oblique niche]

Line 𝑙 belongs to a directly proportional relationship.
For 𝑥 = 9, 𝑦 = 6.
The line 𝑘 passes through the points 𝐴(−2,2)
and 𝐵(2,3).
Calculate the coordinates of the intersection of
the lines 𝑙 and 𝑘.

[oblique niche]

Write the equation of the line 𝑛 through the
point 𝐶(−5,3) which is parallel to the line
𝑙: 2𝑥 − 5𝑦 = 8

[tidal hawk]

hii

so find the gradient for the line l

then that’s the whole eqn

find the gradient for k and use y-y1=m(x-x1) for its eqn

then equate both eqns and ya

[tidal hawk]

since it’s parallel, gradient is the same.

so you have y=mx+c with m known

so just plug in (-5,3) and solve for c

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[plain plaza]

i need help with these (and some others) really badly

[rich roost]

Do you know the properties of logarithms?

[plain plaza]

no i completely forget how to do any of these algebra type ones

[rich roost]

So these are the ones you need for most of this exercice

Your goal is to separate the big term inside the logarithm into smaller ones

Can you try number 2? It’s easier to help if there’s something more specific to point out

I can also just do one of the problems you want and explain it step by step

[plain plaza]

yes pls

im so confused

is #3 right for 2

that one i think i got the rest are ????

[rich roost]

Yes #3 is right for 2)

This is the step by step process for it

For 3), try to factor out 36

Express 36 as a product of prime numbers

[plain plaza]

ooh

[rich roost]

And then 4) is using the same tricks for 2)

Except you replace the values you get with the equations given once it’s simplified

[plain plaza]

is 2 right for 3

[plain plaza]

aaa complicated

[rich roost]

Yeah a bit

[rich roost]

It’s not quite right, I think you forgot to do one more step

Don’t forget to take the exponents out of the logarithm

I think it’s because you did this mistake at the end

You thought the 2 were equal

[plain plaza]

hm

[rich roost]

Anyway take your time to finish them and also send pictures of how you did the exercices, I’ll respond in a bit tho cause I have to go

[plain plaza]

i still have no idea how to do these

[rich roost]

Can you show me what you did for 3)

[plain plaza]

i just factored it so it’s like 2x2x3x3=36

[rich roost]

Right and it’s inside a logarithm

So can you separate these terms to get either log(2) or log(3)

From log((2^2)*(3^2))

[plain plaza]

ohh

so which answer do it be

[rich roost]

It’s 1

[plain plaza]

oooh oo

now for the other 6 questions

[rich roost]

Right lol

[plain plaza]

i did a couple

the only thing in this topic i get is the solving type stuff w logarithms and there’s only one waaa

there’s a money question and it’s 3 parts and i think only part b is wrong

i think i should finish page 1 first though

[rich roost]

Yeah try to finish page 1 but I can check for the other questions after

[plain plaza]

yea i dont get these at all

i just wanna go to bed ;__;

[rich roost]

I can give a step by step process for each of them if that’s needed

[plain plaza]

how on earth does this work

[plain plaza]

that would be good

[rich roost]

You have to find x?

Put both sides to the power of 3

And the log cancels out with the 3

[plain plaza]

so (2x-9)^3 and f^3

[rich roost]

I worded it wrong I’ll write it one sec

[plain plaza]

why does 3 go to 5th

[rich roost]

Because you have to do so in order to retain the equality

If a=b, then x^a=x^b

[plain plaza]

but why not 3^5

i mean

why not 5^3

[rich roost]

That’s because it’s not the same operation you apply

To make both sides equal, you have to put the 3 under the exponent or make 3 the exponent

Because 3^5=3^5 and 5^3=5^3, but 3^5 doesn’t equal 5^3

I’m sorry if this isn’t clear btw text isn’t the optimal teaching medium lol

[plain plaza]

lol

but why does one side go to 3 and the other side doesn’t also go to 3

[rich roost]

But it does

It’s because here you cancel the log and the 3

[plain plaza]

ohhhh

[rich roost]

it’s another property

[plain plaza]

so x = 126?

there’s 3 more questions im having trouble with

the 1 and 4 on the front page

this one

and i dont think i did the second part of this one right

[rich roost]

Yes that’s right

[plain plaza]

ok

[rich roost]

So for #1, here’s the graph for log(x)

Do you notice anything

If not it’s fine I can just explain what’s going on

[plain plaza]

yes

there’s nothing that hits 0

i learned abt that

[rich roost]

Yes exactly, and then in the negatives there is no value for this function

This means the domain is ]0, infinity[

[plain plaza]

so how do i know what answers right

calculator says they all fit

[rich roost]

Now in this question, you have to see what’s inside the logarithm and look for when it’s above 0

[rich roost]

Can you send a screenshot of that

Cause 3 of those don’t work if you’re not working with complex numbers

[plain plaza]

ohh they have to be whole numbs

now i dont see any of em

[rich roost]

Yeah anything above 0 works

So for which value of x does x-3>0

[plain plaza]

so how do answer

[rich roost]

Is the question you need to answer

[plain plaza]

4

[rich roost]

Yeah that’s it

[plain plaza]

i figured out the stomach virus one i think

[rich roost]

So that’s how you need to solve this problem, find what is inside the logarithm and find when it’s above 0

[plain plaza]

okiedoke

i just need 4 and to fix the other one

[rich roost]

Also complex numbers aren’t things like 1/2 its something else but don’t mind it

Alright let’s do 4 then

Try to separate the terms

Inside logarithms, you can split terms that multiply or divide

[plain plaza]

ok ?

[rich roost]

So here you’re able to split the x, the y and the z into 3 different logarithms

log(a/b)=log(a)-log(b)

So this allows you to split the z in this case

log(((x^2)y)/root(z))=log((x^2)y)-log(root(z))

[plain plaza]

weird

[rich roost]

Basically here replace x and y by anything you want

And the equations will stay true

You just have to know them to solve these problems there’s nothing else you can do

[plain plaza]

i just wanna go to beddddd

what would the answer be for 4