help-10

[wet flume]

is it a?

[pine sail]

The answer?

[wet flume]

2/3

[pine sail]

Close.

[wet flume]

-2/3?

[lone echo]

start simpler

also, dont answer randomly

[wet flume]

hehehe

[lone echo]

do you know the limit of sinx/x?

[wet flume]

1

[pine sail]

He's evaluated the limit already.

[lone echo]

didnt see that, where is that?

[pine sail]

.

[wet flume]

@lone echo

[lone echo]

okay. So you know the definition of continuity means that the limit exists, the function exists, and both are equal, correct?

[wet flume]

ok

[lone echo]

the limit exists. It's 4/9.
The function exists. It's k^2
Both are equal. k^2 = 4/9.
Solve for k.

[wet flume]

oh so it's +-2/3

so we have to just equate both..

[lone echo]

that's the definition

[wet flume]

oh

the definition was complex

[lone echo]

limf(a)=f(a)

[wet flume]

for me to understand

[lone echo]

was it?

[wet flume]

m english not so proper

[lone echo]

fair enough

[wet flume]

Thanks a lot @lone echo @pine sail

.close

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[umbral rose]

,tex Could some one help me with finding a Borel subset $A$ of the real numbers such that for every bounded open interval $I$ it follows
$ 0 < \lambda(I \cap A) < \lambda(I) $
I tried to consider $A$ are the irrational numbers, you get almost that but the last < is equality, I could not get any further.

[warm shale]

jfab

[obtuse pebble]

@umbral rose Has your question been resolved?

[umbral rose]

<@&286206848099549185>

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@umbral rose Has your question been resolved?

[obtuse pebble]

@umbral rose Has your question been resolved?

[obtuse pebble]

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[naive grotto]

i need help

[naive grotto]

thats what i have#

im stuck however

idk what to do now

no oje wanna help

😦

😞

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[mint anvil]

Hi community, I'm wondering how to tell what type a knot is given it's crossings

[mint anvil]

I have a knot that I counted 8 crossings for. But I see there are 8_1, 8_2 ect. variants. How could I know which variant I have?

[obtuse pebble]

@mint anvil Has your question been resolved?

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@mint anvil Has your question been resolved?

[obtuse pebble]

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[lean ingot]

Is this correct? The answer is supposed to be m = 1/2, I am supposed to find the slope.

[foggy jasper]

please type the question

[daring rock]

No, that's right, the slope is -2

unless you wrote the equation wrong. Are you sure you have everything in the right places?

[civic zealot]

the slope of that line or the slope of a perpendicular line?

[lean ingot]

4x + 2y = 16, find the slope (m)

[lean ingot]

the line, not the slope of a perpendicular line

[lean ingot]

Yes I have everything in the right places, the answer sheet may be incorrect.

[foggy jasper]

the slope is -2

[lean ingot]

Thank you, I think I will look into the possibilities of error in the answer sheet after gathering more information on the topic.

I will close the channel now

.close

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[plucky sapphire]

I need help with the fourth question please. I don't really understand which way I'm supposed to match them

[tardy epoch]

Just match highest acceptable price the with person who has a maximum willing to pay price one by one

Chad has the highest acceptable price. Match him with Bashir since he's willing to pay that price

[plucky sapphire]

oh ok

Would I combine them after that?

[tardy epoch]

Oh sorry it says largest surplus

I gave you the answer to minimum surplus

[plucky sapphire]

huhhhhhhh

😭

[tardy epoch]

I think you do the reverse of this

Hmm no that's not it

Sorry I don't know anymore

Feel free to ping helpers

[plucky sapphire]

ok

[plucky sapphire]

<@&286206848099549185> I need some assistance please

[humble moss]

They want you to match up the buyers and sellers in such a way that the total surplus (adding the differences between the maximum pay and minimum price) is at the highest possible.

I'm pretty sure they mean the average though

It's weirdly worded

[plucky sapphire]

hm ok

I'll try that

[humble moss]

How you pair it up shouldn't affect the sum of the surpluses though

[plucky sapphire]

I keep getting 25 but it says that's wrong

[humble moss]

Try 26

(17-6) + (14-7) + (13-8) + (12-9) + (11-10) + (10-11) = 11 + 7 + 5 + 3 + 1 - 1 = 18 + 8 = 26

That should work because it doesn't matter how you pair it up. The surplus total will always be the same no matter how you pair it

@plucky sapphire Did 26 work?

[plucky sapphire]

no

[humble moss]

😭

How you pair it shouldn't matter at all if it's a sum of the surpluses

[plucky sapphire]

it was 27

[humble moss]

lol

[plucky sapphire]

It has been done

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[stable willow]

y'(x) = (3 y(x)^2)/(3 y(x) + 14) - (45 y(x))/(3 y(x) + 14) - 14 y(x)

Can any1 help with this problem

[stable willow]

This equation is what I've derived, my goal is to find y(x)

<@&286206848099549185>

[dawn dock]

y(x) or y'(x)?

[stable willow]

This is the equation for y'(x) I want to find y(x)

[obtuse pebble]

@stable willow Has your question been resolved?

[earnest galleon]

hey

[stable willow]

Hi

[obtuse pebble]

@stable willow Has your question been resolved?

[obtuse pebble]

@stable willow Has your question been resolved?

[stable willow]

why is nobody answering :(

<@&286206848099549185>

[brazen viper]

@stable willow This should be a separable equation. Use y'(x) = dy/dx then move all the terms with y to the side with the dy/dx, then integrate both sides with respect to x (the dy/dx will "cancel" with the dx to effectively create an integral in terms of y). Finally, solve the resulting equation for y. Don't forget the +C.

[obtuse pebble]

@stable willow Has your question been resolved?

[stable willow]

alright I'll try later

[obtuse pebble]

@stable willow Has your question been resolved?

[echo steppe]

@stable willow

[stable willow]

yes.

you know what I should probably go learn integration while I'm at it

If I get stuck I'll ask here again

adios

.close

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[harsh goblet]

Hi! How could I make a function that takes in a math equation and a given variable that is in the equation and solves the equation to that given letter?

[harsh goblet]

do you know any like free api of sg?

[obtuse pebble]

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[cloud radish]

Discord question. How do I shut off the bot quoting this server?

[last pilot]

?

[cloud radish]

I have a running cometary coming from this server. How do I turn it off?

[last pilot]

i have no idea what you mean

[devout yarrow]

dont ask it here

[cloud radish]

recompile 321 say etc

stuff like that

Where do I ask?

[devout yarrow]

oh nvm

what r u asking abt exactly

[cloud radish]

Turning off this server from giving a verbatim sound file of everything posted.

[last pilot]

...
it doesn't do that?

bro

u have text to speech on

XD

[cloud radish]

Ok. How do I stop it?

[last pilot]

google

.close this channel

type .close

[cloud radish]

.close

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[cyan mason]

Help solving this integral

[versed stratus]

I'd probably go with IBP

[cyan mason]

Doesnt work

[cyan mason]

I tried

[versed stratus]

hmm, have you tried x=e^y then

[cyan mason]

It keeps on increasing

No

[cyan mason]

Can you solve it please

[versed stratus]

Let me try, I'm not sure if I can. Mind if I check if its even integrable with wolfram first?

[cyan mason]

Sure you can

[versed stratus]

,w integrate (1+ln(x))^2/(1+x^2)

[cyan mason]

Because I have tried solving myself

[warm shale]

Results provided by WolframAlpha

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

Requested by physicsrocks

[versed stratus]

yup,, this is beyond me, sorry

I don't know $Li_3 (x)$

[warm shale]

Why am. I here

[versed stratus]

,w integrate e^x(1+x^2)/(1+e^2x)

[warm shale]

Results provided by WolframAlpha

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Requested by physicsrocks

[versed stratus]

ok, this is beyond me, I'm very sorry

[cyan mason]

Ok no problem

Thanks tho

It's out of my syllabus

So no worries

I thought it was easy

[versed stratus]

did your teacher give you this problem?

[cyan mason]

Kind of just to check if it's even correct

[obtuse pebble]

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[timid silo]

How can I solve it

[timid silo]

I tried substituting value of x from eqn. 1 in eqn. 2 but it doesn't matches with the options

[leaden lion]

Show ur work

[torpid cave]

What can you solve for that is of the form $ax + by$

[warm shale]

casework

[timid silo]

I am getting the value of $x$ as $x=(1-qy)/p$ from first eqn

[warm shale]

Singularitrix

[torpid cave]

No dont do that

[timid silo]

so then

[torpid cave]

Ignore what is x and what is y

You want x + y

[timid silo]

yes

[torpid cave]

So what do you want in the end?

You want something of the form$ (x + y)(\text{something}) = \text{something else}$

[timid silo]

yes

[torpid cave]

How can you achieve that

[timid silo]

but actually , I think elemination will not do it anyway

[torpid cave]

First get everything of the form ax + by on one side and everything else on the other for both equations

[timid silo]

Ok sure, i will try

thanks

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[timid silo]

if f and g are functions why does it follow that (fg) (x) = f(x) g(x)

[timid silo]

?

[sage geode]

That's the way fg is defined in this context

[timid silo]

how so

[sage geode]

You just replied to the answer

[timid silo]

does it follow that fg = gf as well

oh right

[sage geode]

If I defined h(x) = x^2, what's the point of questioning it?

[sage geode]

Yeah

[timid silo]

ah i think i see it

thxthx

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[gritty swallow]

how is this not correct? for part c

[gritty swallow]

the function that we got from solving the difeq is 45 + 45 e^(-0.007 x) which is right

nevermind

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[compact spruce]

Can someone explain q5 part c

[compact spruce]

??

[unique nebula]

!show

[obtuse pebble]

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

[compact spruce]

It's a function quest8on

Question

[obtuse pebble]

@compact spruce Has your question been resolved?

[obtuse pebble]

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[warm shale]

kisnar

[gilded needle]

log is monotonically increasing

so it doesn't change the inequality direction

x < y <==> ln(x) < ln(y)

[tardy tusk]

it's ln

[gilded needle]

yeah log of any base i meant

they're all related to each other by scale factors

[tardy tusk]

but if log 0 to 1 then flip

[gilded needle]

wha

there is no log 0

[tardy tusk]

(0;1)

[gilded needle]

no the inequality does not flip there either

[warm shale]

kisnar

[tardy tusk]

Hmm, strange

[gilded needle]

log itself goes negative for x < 0 but it's still a monotonically increasing function there

[tardy tusk]

Well you can say that 0 is ln(something)

[gilded needle]

yea, the function f(x) = e^x is also monotonically increasing

so you can exponentiate both sides without changing the inequality

[warm shale]

kisnar

kisnar

[gilded needle]

yep

[obtuse pebble]

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[cedar fog]

How do you find the domain?

[cedar fog]

How to find domain?

[tardy tusk]

You sent it

[cedar fog]

Dang, why can't I send it here 😭

[tardy tusk]

Three times

Picture

[cedar fog]

oh

I can see it now.

great

[obtuse pebble]

@cedar fog Has your question been resolved?

[obtuse pebble]

@cedar fog Has your question been resolved?

[obtuse pebble]

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[arctic heron]

i figured out the period of cos2theta by doing 2pi/n and i can visualise the period of cos^2theta but i don't know how to show it

[torpid cave]

Can you show period of $|\cos\theta|$

[warm shale]

casework

[arctic heron]

period of costheta would be 2pi/1

oh mod costheta

i pretty sure its just half of 2pi cause the negatives flip up

but thats the only way i can think of calculating it

[obtuse pebble]

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[torpid cave]

Any ideas?

[merry wave]

well

not really

[torpid cave]

When will $6^y$ not be a factor of $2^{14}3^{24}$

[warm shale]

casework

[merry wave]

bc i can do it the long way which is 2x2x2 and so on until i reach 14 times and same concept with the 3 and do the same concept with 6^y but tbh thats a long time and im pretty sure there's a shortcut lowkey

[merry wave]

not sure

[high lily]

consider exponent laws, and you won't have to write out a large bunch of 2s and 3s

[merry wave]

arent you adding the exponents?

or no?

[timid silo]

break 6 up into 2*3

[merry wave]

ok

[torpid cave]

Or turn $2^{14}\cdot 3^{24}$ in something of the form $6^x \cdot k$ both viable options

[warm shale]

casework

[merry wave]

sorry im confuse

i think you multiply exponents

[torpid cave]

More like all these

[merry wave]

YEAH I KNOW THOSE

oops didnrt mean to write in caps

[torpid cave]

You are looking for power of a product rule

[merry wave]

yeah so 6^38

i figured that out

@torpid cave

[torpid cave]

Ummm....

You sure that is right?

[merry wave]

wait

power of a product rule

what is that

[merry wave]

so (3)(2)=6

but what is m

we have two exponents

[timid silo]

m is just whatever is in the exponent

[merry wave]

no we have 2^14 and 3^24

[timid silo]

in this case, y is in the exponent

[merry wave]

i know

but its split

how do we put it back together

do we multiply the exponents

[timid silo]

6^y is a factor of 2^14 * 3^24, but 2^14 * 3^24 isn't necessarily expressible as 6 raised to some power

the goal isn't really to combine 2^14 and 3^24

[merry wave]

ok

[timid silo]

knowing that 6 equals 2*3, what does this rule tell you about 6^y?

[merry wave]

idk?

[timid silo]

a is 2, b is 3, m is y

[merry wave]

yes

m??

ah i see

you're referencing off the image

i know that

but idk how to find 'm'

[timid silo]

i literally just said it

[merry wave]

you literally just said

m=y

that doesnt say much

[timid silo]

ah

[merry wave]

how do i find the greatest potential value of 'm'

because all i know now is 2*3=6

but idk anything about the exponents

[timid silo]

ok, does it make sense that 6^y is 2^y * 3^y

[merry wave]

yes

[timid silo]

sorry, having trouble with phrasing

[merry wave]

ok

[timid silo]

well, in 2^14 * 3^24, there are more 3 factors than 2 factors

but in 6^y, they're the same

meaning the value of y is capped by the amount of 2 factors

[merry wave]

ok how about this

provide me an example

a different example so i can understand it

explain that example and maybe that will help

[timid silo]

ok, let's just use lower numbers than 14 and 24

say, 2 and 4

[merry wave]

ok'

[timid silo]

6^y will be a factor of 2^2 * 3^4

[merry wave]

ok

[timid silo]

2^2 * 3^4 is just 2*2*3*3*3* 3 (no need to evaluate really, so i'll leave it like that)

[merry wave]

ok

[timid silo]

whenever you raised 6 to a power y, it can be shown in the form 2*3 * 2*3 * 2*3... y times

oops

[timid silo]

the amount of factors of 2 that 6^y has will be the same as that power y

[merry wave]

wait

[merry wave]

explain this again

wdym 2*3 y times

we dont know how much y is

[timid silo]

well no, but we know that whatever it is, 6^y is equal to 6 times itself y times, which is equal to (2*3) times itself y times

[merry wave]

arent they both the same

[timid silo]

yes

[merry wave]

2*3 and 6 lol

[timid silo]

yes

[merry wave]

ok

[timid silo]

but splitting it up shows the factors a bit more explicitly

[merry wave]

ok

[timid silo]

so, when you raise 6 to some y, there's y (2*3) terms multiplied by each other

which means there's y 2 terms and y 3 terms

[merry wave]

ok

[timid silo]

in this smaller example, there's 2 2 terms and 4 3 terms multiplied be each other

[merry wave]

ok

[timid silo]

so since 6^y is a factor of 2^2 * 3^4, the amount of 2 terms and the amount of 3 terms in 6^y has to be less than or equal to the amount in 2^2 * 3^4

in other words, y<=2 and y<=4

the greatest possible y which satisfies this is 2

[merry wave]

why is it 2

oh i see

so in my example its y<=14 and y<=24

[timid silo]

yes!

[merry wave]

so what would my answer be?

[timid silo]

the greatest possible y which satisfies both of those inequalities is

[merry wave]

none of these satisfy

i mean

wait

yeha no i dont see anyone satifying

[timid silo]

your conditions are y<=14 and y<=24

[merry wave]

yes

[timid silo]

any number <=14 sastisfies both of those inequalities

for example, 7

[merry wave]

yes

[timid silo]

7 is <= 14 and 7<= 24

[merry wave]

oh i see

so 14 is the answer

[timid silo]

yea

[merry wave]

ty

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[haughty shore]

for b why can i not treat it like a+b+c+d+e+f=20 but everything but a cant be 0 and then it would just be 20 choose 5

but thats not correct

[fossil crag]

definitely not a+b+c+d+e+f=20

I can always pick 10 12 14 16 18 20 for example

I doubt the sum of those is 20

or are you referring to something else?

[haughty shore]

oh bruh i cant read lmao

ty

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[fathom flicker]

Could anyone help me understand this notation in my book? I can't make sense of what this vector field is meant to be

[fossil crag]

$F(x,y,z) = (x^2+y^2+z^2)\cdot \begin{pmatrix}x\y\z\end{pmatrix}$

[warm shale]

rafilou2003

[fossil crag]

(i,j,k) is the canonical basis of R^3

[fathom flicker]

so

F=(x^3+y^2+z^2, x^2+y^3+z^2, x^2+y^2+z^3)?

[fossil crag]

that's not how multiplication works

(x^2+y^2+z^2)x = ...

and then for the other 2 coordinates

[fathom flicker]

oh right

ty

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[fathom flicker]

glad you're the only one catching all my dubious mistakes tnight

[fossil crag]

It's 4AM for me, I'm the math vigilante

[dark stirrup]

I'm a mistake making machine if I try to help at such a late time

[obtuse pebble]

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[mint nebula]

For how many integer values of x, the expression is <= 0

So for the expression to be negative, we need to have an odd number of negative factors.

x can be 0,2,3,4 from what ive found, but I don't seem to see a rule in calculating the number of values,

I dont need to know the exact values, just how many are them.

[obtuse pebble]

@mint nebula Has your question been resolved?

[mint nebula]

<@&286206848099549185>

[daring sorrel]

0 does not work, 1 does work

Something you may notice is if x is a perfect square then it works because one of the numbers will be 0. Try considering the values between perfect squares

If x is between 99^2+1 and 100^2 it will work, because 99 of the numbers will be negative, but if x is between 98^2+1 and 99^2-1, 98 of the numbers will be negative, hence it is now positive and doesnt work

[mint nebula]

Yes, my mistake, 0 doesnt work.

So like

We have 100 perfect squares, so 100 values

But then

[mint nebula]

I didnt quite get how to calculate the number of x values from this

[daring sorrel]

Well do you see the pattern

[mint nebula]

Not really

[daring sorrel]

Alright I will try explaining it more

Let us start at x=0. First you should try and reason to yourself that whenever x is negative, the entire product will be positive, so we can ignore those.

Next, what happens to the signs of each (x-a)^2 as x increases? Starting from the left, as x gets bigger the terms become negative

[mint nebula]

Indeed.

[daring sorrel]

Now this change happens everytime x=a^2 for some a from 1 to 100

[mint nebula]

So everytime x is a perfect square, one more (x-a) changes signs

[daring sorrel]

Yeah

In between x=a^2 and x=(a+1)^2, either all of these numbers work, or none of them work

[mint nebula]

mhm

[daring sorrel]

So thats the main pattern. Now you just need to figure out how to count all those values

Remembering the connection between perfect squares and the sequence 2n+1 will help you count them

[daring sorrel]

Btw to be more specific, this switches between every other perfect square

So between 1^2 to 2^2, 3^2 to 4^2, 5^2 to 6^2, … all numbers in between are negative, for 0^2 to 1^2, 2^2 to 3^2, 4^2 to 5^2, … all numbers in between are positive

The distance between a^2 and (a+1)^2 is 2a+1, so you will count them by considering all odd values of a

[mint nebula]

So from odd square to even square we have negatives

And from even square to odd quare we have positives

Thats gonna take ages tho

[daring sorrel]

Well with a calculator it shouldnt

There are gonna be 50 “things” you are adding together if you do it right, and you can also just express it in summation notation

[mint nebula]

I dont think my calculator can handle that

Ill try the google one

[daring sorrel]

Also there is a formula for it too

You can use sum of first n natural numbers = n(n+1)/2

[mint nebula]

Maybe i can try that

[mint nebula]

Oh, that gauss' sum

But with what numbers do i use it here

So like

The numbers in between the odd and even squares

[daring sorrel]

Yeah

There will be 2(odd number)+1 of them

[mint nebula]

Bht why the sum

I know the formula (last term-first term)+1

[daring sorrel]

I dont understand

[mint nebula]

Me neither.

[daring sorrel]

Let me make sure we are on the same page, if I asked how many numbers are between 55^2 and 56^2 could you tell me (w/o using calc)

[daring sorrel]

Actually I guess this is right

But you want a simpler formula that is easier to compute

[daring sorrel]

Which is given by this

[mint nebula]

112

[daring sorrel]

Yeah

[mint nebula]

Alright so now i gotta calculate using this

To all squares

Then add them up

Right?

[daring sorrel]

Yeah

[mint nebula]

Okay let me do that rq

[daring sorrel]

But only for the odd to even

[mint nebula]

Yeah, i know, of course

@daring sorrel ive gotten 5050

[daring sorrel]

What did you do to get this I am getting something different

[daring sorrel]

What is this

[mint nebula]

this

Oh, crap

i've made a mistake, sorry

it's 199+200*50/2

it's 9975

[potent burrow]

Heloe I want help is there formula to find whether two lines intersect or not

[daring sorrel]

lol im confused now im getting 3+7+11+...+199=5050 which should be correct

I think you were originally correct. There is one last thing though we need to account for

The 11 for numbers between 5^2 and 6^2 is only counting 5^2 to 6^2-1, but we also know even squares work, so we need to add 1 for every possible even square (which there are 50 of)

so in total you get 5050+50 = 5100

[mint nebula]

Right.

.close

[obtuse pebble]

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[cunning flower]

The last exercise last question

<@&268886789983436800> someone understand French to help pls

<@&286206848099549185>

.close

[obtuse pebble]

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[jade mural]

guys hello, why did he pick N-0 as max interval ?

[kind hawk]

well in the argument you want that both n>=9A and that n>=2

so you can let n be the maximum of those

I dont know what the E means tho

rounded up maybe?

[jade mural]

E means round yes

but backwards

like -1.1 to -2

i think

because

make it an integer

to use arcemidian(typo)

as there exist a natural number always bigger than

n_0

to make the implication hold

the thing is why i need the max

[kind hawk]

.

[jade mural]

hmm

let me see if i understand

wo

ow so i know that n-0 need to be greater than

9A

so we get the integer of 9A by E(9A) +1

so its bigger

but how the 2nd argument came ?

I didnt get that one

The n^2 -1 part

[kind hawk]

thats not in the max

[jade mural]

max is ...., 2)

2 is in there right ?

[kind hawk]

you want to use that n^2-1 >= n^2 / 3

which is just false for n=1

so you need n>=2

[jade mural]

last Q

is the max an interval here ?\

[kind hawk]

max of two numbers is a number

[jade mural]

ow like the max between them ?

interesting

[kind hawk]

ok were you from the start confused about what the notation actually means?

max{a,b} means the bigger of a and b

[jade mural]

u mean one of them ?

the bigger

gets chosen ?

[kind hawk]

max{5,9}=9

[jade mural]

ok i didnt know that

i thought its an interval

thank u for that

i see now

thank u @kind hawk

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[timber pine]

how do i do this question?

[timber pine]

i know concepts such as F=mgsintheta and costheta

i tried to let the force thats sloping downwards be mgsintheta, which gives 68.8N but the answer is Wcos55 = 675 N

[obtuse pebble]

@timber pine Has your question been resolved?

[obtuse pebble]

@timber pine Has your question been resolved?

[fleet ridge]

@timber pine yes this is true

But what exactly is it asking us

maximum friction?

[timber pine]

ye

[fleet ridge]

one minute

force perpendicular to the surface is cos35°

×120N

right?

Well if it needs to slide, the friction force must be smaller

68.8N-k•120N•cos(35°) > 0

shouldn't it be like this?

or am i wrong

Friction Force = kN

k is coefficient of friction

[timber pine]

i think?

but the ans is Wcos55

[fleet ridge]

what is W

okay

weight?

But i think it won't move if it's equal

But if we want it to be equal, this is it, yes

[obtuse pebble]

@timber pine Has your question been resolved?

[timber pine]

why tho

[fleet ridge]

horizontal force 120N×cos55 or 120N×sin35

because of that

Since the friction force will be opposite

[timber pine]

which

[fleet ridge]

hypotenuse of triangle

So I'm talking about sloping ground

like this /

When we divide the weight into its components

120N×sin35 or 120N×cos55

and,

120N×sin55 or 120N×cos35

[fleet ridge]

The forces perpendicular to the hypotenuse of the triangle are as follows 120N×sin55 or 120N×cos35

both the same

Forces making an angle of 0° with the hypotenuse are 120N×sin35 or 120N×cos55

both the same

[fleet ridge]

If we want the friction force to be maximum we should set this equal to one of these two

because when it is equal to one of these it becomes maximum, if it is within its capacity

But apparently the friction coefficient is suitable for this, otherwise I think he wouldn't have asked

or i missed something

whatever do you understand?

In other words, if the coefficient of friction was very low, this object would slide, but if it was something large like 0.7 according to my calculation, it would be enough for it to stop

[timber pine]

I do not think that is involved in this question at all

[fleet ridge]

but you said that was the answer

120N×cos55

Doesn't this mean that this is the maximum friction?

at least for this object

[timber pine]

Wait

hold on

[fleet ridge]

ok

[timber pine]

ans says

F=Wcos55 = 120x9.81xcos55 = 675N

[fleet ridge]

mg²?

[timber pine]

I dont know

[fleet ridge]

what

this is

[timber pine]

perhaps a pic woul help?

[fleet ridge]

maybe

let me look

[timber pine]

q6

could it be a printing error

[fleet ridge]

yes

I think so too

[timber pine]

what'd you get?

[fleet ridge]

120N/9.81×9.81 should have been

because says W

Isn't this Weight?

We don't have a problem with cos anyway

Weight = 120N must be like that

[timber pine]

I think i understand why u said sin35 or cos55 is right since sin(90-x)=cosx

[fleet ridge]

but too much it is

[timber pine]

yup

[fleet ridge]

yes

[timber pine]

hmm

[timber pine]

thanks for ur help

i'll ask my teacher about it

[fleet ridge]

okay

[timber pine]

ty

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[fierce jolt]

Help

[inland valley]

💀

[fierce jolt]

I'm serious

[inland valley]

what grade is this

[fierce jolt]

Please help

[fierce jolt]

Idk it's a quiz

There's a thing in my singing monsters where you get diamonds from other games

I'm trying to get like 282 coins

JUST HELP ME

[inland valley]

lmao

right

alr im just subbing values

[fierce jolt]

So 🌈+2=🦄

So
1+3+3≠25
2+4+4≠25

Check everything

4+6+6=16≠25

6+8+8=22

7+9+9=25

🌈=7

🦄=9

Let's change ❤️ to x

X+7+7=35

35-14=X

21

25-21=4

.close

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[vast willow]

you can find the solution without trying random numbers

[obtuse pebble]

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[bold tartan]

This is a question from sample 1 paper of the IC aero MAT test (which i am sitting next next monday and still cant get things like this right so im kinda cooked)
Please help

[bold tartan]

<@&286206848099549185> ?

[obtuse pebble]

@bold tartan Has your question been resolved?

[tardy epoch]

You can just plug in each of the three options into the differential equation

[obtuse pebble]

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[versed briar]

what terminology would i use to describe the change in frequency with changing tension considering it's under a radical

for example if there was an exponent i could say exponentially

[obtuse pebble]

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[lament slate]

Ok, my turn. Here is a question I am having issues with: (trigonometry)
the specific part i am having trouble with comes down to the first steps of simplifying the fractions themselves.

[crisp zephyr]

The fractions are already simplyfied

[lament slate]

my apolagies my screengrab left out the problem

[crisp zephyr]

tan t = sin t / cos t

[lament slate]

and when i put those fractions over themselves and got to simply- is where im having problems -
3/4/rad 7/4

am i to multiply 3/4 * 4/rad7 ? as the first step

[crisp zephyr]

(3/4)/(sqrt(7)/4) = 3/sqrt(7) = 3 sqrt(7)/7

[lament slate]

ok i just did it on paper and i see who you went from 3/sqrt(7) to 3 sqrt(7)/7, you multiplied the numerator and denominator by sqrt (7) and it gave you 3 sqrt (7)/7 but how did you go from the original fraction to the 3/sqrt(7) ? @crisp zephyr

i figured it out : you multiplied 3/4 * 4/sqrt(7) = 12/4sqrt(7) simplified to 3/sqrt(7) which simplifies to 3sqrt(7)/7

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[terse tapir]

a person buys 5$ worth of eggs from 100 different markets. You have to find average number of eggs per $ purchased from all markets combined
the average usable in this case is

1. arithmetic mean
2. geometric mean
3. harmonic mean
4. none of these

[terse tapir]

hi

i needed help with my stats

[timid silo]

When we say "to average" we add all the prices from all the different markets and then divide by the amount of markets

so

[terse tapir]

arithmetic?

[timid silo]

yes

[terse tapir]

but it is given as harmonic

like

literally everywhere

idk how

[timid silo]

there is no meaning to referring to the harmonic mean here

[terse tapir]

btw there is something relevant to this mentioned in my textbook

it says

when average is to be calculated, of form a/b, where a and b are different quantities then

1. use hm if a is constant
2. use am if b is constant

soo

how is it that in this question the quantity a is constant?

[terse tapir]

for example, average speed is to calculated when distance is given or constant,
therefore since average speed = distance/time and distance (a) is constant, we use harmonic mean

[obtuse pebble]

@terse tapir Has your question been resolved?

[obtuse pebble]

@terse tapir Has your question been resolved?

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[terse tapir]

.reopen

[obtuse pebble]

✅

[tardy epoch]

What does usable mean

[terse tapir]

appropriate average to be used

in the given case

[tardy epoch]

The constant is price

[tardy epoch]

"$5 worth of eggs"

[terse tapir]

yes

[terse tapir]

so if the price is a in a/b
what would be b?

[tardy epoch]

It's determined from reading the question

[terse tapir]

would it be the number of markets?

[tardy epoch]

No

The units are just eggs and $

[obtuse pebble]

@terse tapir Has your question been resolved?

[obtuse pebble]

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[fluid bone]

integer 2~0 (3x+1)^2 dx

[timid silo]

,, \int_2^0 (3x+1)^2 \dd x

[timid silo]

is this it?

[fluid bone]

yeah

so i need to use substitution i know that

but how exactly

i think i turn 3x+1 into t

[abstract flame]

if u=3x+1, then what is du in terms of dx

[fluid bone]

idk whats du

we never used that term

[brisk grove]

He probably hasn't learned substitution

Just expand the parenthesis first

[timid silo]

Fix those integration bounds first

[fluid bone]

2 to the 0

are bounds

other way around

[timid silo]

,,\int_a^b \m fx \dd x = -\int_b^a \m fx \dd x

[timid silo]

you are meant to say that as 0 to 2, then

[fluid bone]

alright then

2 to 0 integer

so whats the first thing i do

[abstract flame]

try expanding out (3x+1)^2

[fluid bone]

a ok so (3x+1)(3x+1) ?

[abstract flame]

fully expand it

with FOIL or whatever it's called

[fluid bone]

is 9x^2+6x+1 good enough?

[frosty crypt]

+1, not +2

[fluid bone]

mb

ok so i expanded it what next

do i multiply with x

?

[frosty crypt]

now just use inverse power rule. on each term

[fluid bone]

i misswrote the assignment hold on

[frosty crypt]

yeah, multiply all the terms by x, then apply inverse power rule.

then use fundamental theorem of calculus to evaluate the definite integral.

[fluid bone]

2 many complicated words sorry xD

what is inverse power rule

[frosty crypt]

just $\int{x^{n}}{dx} = \frac{x^{n+1}}{n+1}$

[warm shale]

fish

[fluid bone]

idk how to apply that in my case

i have interger 2 to 0 (9x^3 +6x +x)dx

[frosty crypt]

well with that, since $\int{f(x)+g(x)}{dx} = \int{f(x){dx}} + \int{g(x){dx}}$, apply that to each term.

[warm shale]

fish

[fluid bone]

i cant think through that

can you explain it to me with my example

ill figure the rule on my own just this way its confusing

[frosty crypt]

okay.

so $\int9x^3+6x^2+x{dx} = \int9x^3{dx}+\int{6x}{dx}+\int{x}{dx} = \frac{9x^4}{4}+3x^2+\frac{x^2}{2}+C$ (ignore the +C for now, unimportant to evaluation)

[warm shale]

fish

[fluid bone]

so i split it all apart

alright

[frosty crypt]

yes

and fundamental theorem of calculus says $\int_a^b{f(x)}{dx} = F(b)-F(a)$ F(x) is the antiderivative. in the case of your problem, it's the $\frac{9x^4}{4}+3x^2+\frac{x^2}{2}+C$

[warm shale]

fish

[fluid bone]

hol up why did you get /4 and /2 in there

explaining it to me with theorems wont work sorry

[frosty crypt]

think of it in reverse, if we take the derivative of both of those, we will get the original functions.

[fluid bone]

ahh ok mb

just integer them okay

[frosty crypt]

yeah

[fluid bone]

but do i need to do it for both 0 and 2?

for each one

[frosty crypt]

just plug in the value for 2 and 0

[fluid bone]

i think you write a line after each one then

with 2 to 0

right

[frosty crypt]

yeah, that's correct notation

[fluid bone]

alright i have formulas for reverse derivation so ill try to come up with it on my own

[frosty crypt]

alright

[fluid bone]

ok i figured it out, all but the middle one

why isnt it 6x^3/3

[frosty crypt]

it should be 6x^x/3, i made a mistake.

[fluid bone]

alright nice

lol

ok so now i input 0 - input 2

for each?

or input 2 - input 0

[frosty crypt]

input2 -input0

[fluid bone]

oki

i got 54 + C is that correct?

i also need help with this one

[frosty crypt]

the C should not be there, it's only used when it's an indefinite integral, otherwise it's correct.

[fluid bone]

okay so after i remove the lines with bounds C disapears?

[frosty crypt]

yeah

[fluid bone]

alright nice

[fluid bone]

what do i do with this tho?

[frosty crypt]

do you know about integration by parts?

[fluid bone]

not really

but when i see it ill understand

[frosty crypt]

but, you did learn about it, correct?

[fluid bone]

i just cant grasp things from theorems

idk we mostly used substitution

but ive heard about it yes

[frosty crypt]

i think integration by parts is required, unless i am forgetting a way to use substitution, wait a bit as i try to find out

[fluid bone]

no its for sure it

try partial integration

[frosty crypt]

yeah, by parts is required, where $\int{f(x)g'(x)}{dx}=f(x)g(x)-\int{f'(x)g(x)}{dx}$

[fluid bone]

shove it to me through my example

[warm shale]

fish

[frosty crypt]

alright

[fluid bone]

theorems just confuse me

[frosty crypt]

let $f(x) = ln(x)$, and $g(x) = x^{-3}$ for this method, you find the antiderivative of $x^{-3}$ first. this gives f(x)g(x), or solved, $-\frac{lnx}{2x^2}$

[warm shale]

fish

[fluid bone]

why to the -3 and not 3?

[frosty crypt]

it's 1/x^3

x^-3 and 1/x^3 are equivalent

[fluid bone]

so antiderivative is 1/f(x)

[frosty crypt]

no

[fluid bone]

or rather through denominator

[frosty crypt]

for the second part, since $\frac{d}{dx}ln(x) = \frac{1}{x}$, you will get$ -\int{\frac{1}{2x(x)}{dx}$, which can be solved with power rule

[fluid bone]

1/denominator?

[warm shale]

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

[fluid bone]

again im confused

so for g(x) i need antiderivation

[frosty crypt]

yes

[fluid bone]

which is 1/denominator

or numerator

depending on which i choose

but generally f(x) i numerator and g(x) denominator

[frosty crypt]

not exactly, it depends on which is easier to find the antiderivative.

[fluid bone]

but in every g(x) i need to do the 1/selected denominator/numerator

[frosty crypt]

well here, g(x) is $-\frac{1}{2x^2}$

[warm shale]

fish

[fluid bone]

thats derivative of x^-3?

[frosty crypt]

antiderivative

[fluid bone]

or rather anti

yeah

[frosty crypt]

yes.

[fluid bone]

why isnt it x^-2/2

like the first example

[frosty crypt]

the negative comes from the -3 in the exponent, so it is x^(-3+1)/(-3+1), so x^-2/-2

[fluid bone]

okay sorry my brain is fried for today with math ive been doing it for too long but thank you for your help

[frosty crypt]

no problem

[fluid bone]

.close

[obtuse pebble]

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[warm shale]

kisnar

kisnar

[torpid cave]

All numbers on LHS are positive

Not real ones

Then yeah no sols

Not everyone sees it though

[obtuse pebble]

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[obtuse pebble]

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[arctic escarp]

Can someone help me with this assignment

[arctic escarp]

Yes.

@timid silo are u helping?

[timid silo]

No

[arctic escarp]

ohk

<@&286206848099549185>

[jolly fossil]

Hi

[arctic escarp]

HI

[jolly fossil]

Whats the problem mate

[arctic escarp]

I just dont understand

and i need help from 1-7

[jolly fossil]

Okay so you know how to add vectors yeah?

[arctic escarp]

no

[jolly fossil]

Forget about dot product

[arctic escarp]

maybe

[jolly fossil]

So vectors in 2d are just 2 numbers
One of them is for x
And one of them is for y
When we say we have
2 x^ + 3 y^
It means the vector represents going 2 blocks in x hat
And 3 blocks in y hat

[arctic escarp]

Ok

[jolly fossil]

If we want to add to vectors to each other
We add numbers in x hat with each other and numbers in y hat with each other
For example
A=c x^ + d y^
B= f x^ + z y^
A+B = (c+f) x^ + (d+z) y^

And for
A-B
We just right
A+(-B)
And -B is B with those numbers having a minus
For example
B= f x^ + z y^
-B = -f x^ -z y^

[arctic escarp]

Ok

So

How do we start with #1

[jolly fossil]

For the first problem
We want to find AB
Lets call it X
We know if we add X to A
we Will get B
So
A + X = B
arranging this
B - A = X (these are vectors)
So you need to subtract x hats and y hats individually

[arctic escarp]

I’ll brb

Give me 10 mins

[jolly fossil]

A = 3 x^ + 7 y^
B = 8 x^ + 4 y^
B - A = (8-3) x^ + (4-7) y^
X = 5 x^ - 3 y^

Take your time mate

[obtuse pebble]

@arctic escarp Has your question been resolved?

[arctic escarp]

@jolly fossil check msg reqs

[obtuse pebble]

@arctic escarp Has your question been resolved?

[obtuse pebble]

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[bold bane]

What have you tried?

[wet flume]

i don't know how to approach

[bold bane]

So you have a garden that is in the shape of a square and a circle.

You have a fence that is 28 meters long that you want to wrap around both gardens.

The question states the circumference around the circle garden is x units in length.

If its circumference is x units long, what is the circles radius?

[obtuse pebble]

@wet flume Has your question been resolved?

[obtuse pebble]

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[astral ivy]

Can a polygon always be chopped into pieces and reassembled to form a square?

[fathom flicker]

rational side length?

[astral ivy]

That has no impact on the problem

Also: finite number of pieces, no non-measurable sets or other bs

Just a simple geometric question

[shell creek]

What are you even asking for tho?

[astral ivy]

Wdym? I’m asking if polygons can always be chopped into pieces and reassembled into a square

[shell creek]

What definition of polygon are we using here?

[cobalt shuttle]

think you need to narrow down a lot more what chopped into pieces and reassembled means
if you only allow rigid transforms, and all edges have to stay edges (no putting two edges together to cancel out) then you need your original shape to have 90* angles, anything else can't be "bent" by a rigid transform