help-13

[livid hound]

directly applicable here

[rancid wasp]

So the answer would be the 38?

[livid hound]

yes

[rancid wasp]

Okay thank you

.close

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[vagrant halo]

where did 0.01 come from

given that dm/dt =QinCin - QoutCout

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[crimson sedge]

Can someone help me find a topic?

[cedar kiln]

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[umbral dew]

bruh

[crimson sedge]

.reopen

Can someone help me find a topic?

[umbral dew]

it's not gonna open so pls open a new channel

[crimson sedge]

whelp me.

[umbral dew]

with what?

open another channel just how u opened this one

[crimson sedge]

Uh.

[crimson sedge]

Are you a helper?

I am looking too, but there is no one avalibale.

[cedar kiln]

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[glad barn]

Can someone explain how this example is supposed to make sense

[glad barn]

Definition 1.3.1
The sentences of propositional logic language L based on the non-empty set of atomic sentences P are formed as follows: 
1. Every atomic sentence p∈P is a sentence of the language L. 
2. If α and β are sentences, then (¬α), (α∨β), (α∧β), (α→β), (α↔β) are also sentences of language L. 
3. Only strings formed on the basis of the two previous rules are sentences of the language L. 

Example 1.3.2 If P={p,q} , then for example p,q,(¬p),((¬p)∨q) and (((¬p)∨q)→p) are sentences of language L, but not (¬()) nor (p∨r).

How is (p∨r) not in the language? p and q are in the language. Then by (2.), (p∨r) should be in the language. Well sure, maybe they mean remove the outer parentheses. But if that were the case, (((¬p)∨q)→p) and ((¬p)∨q) wouldn't be in the language.

[dusk finch]

r isnt in the set P

P = {p, q}

[glad barn]

Ah

[dusk finch]

that's the set of atomic sentences by definition

so not even "r" is a sentence

[glad barn]

Ait yeah, I missed that. Cheers.

!close

[dusk finch]

.cloise

.clsoe

.close

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[dapper mantle]

Hello, I am doing my Matrix homework. I am doing the next problem where I have to solve for X. And I am stucked in this part:
(A . X . B)^t
(None of the matrices are symmetrical)
I am trying to apply the next property:

• (A . B)^t = B^t . A^t
  But I don't know if it's the correct solution.

[rigid lava]

,rotate

[rigid lava]

use the property that det(M) = det(M^T)

so in the bottom line, take the det of both sides

[dapper mantle]

So the det of (A.X.B)^t is the same as the det of A.X.B

• | A.X.B | = |2.C.A|

[rigid lava]

yes

[dapper mantle]

Thanks, I forgot about that one

[rigid lava]

np

[dapper mantle]

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[robust fulcrum]

hi can someone help me get the radius of the circle inscribed within a sector with upper angle 60

radius of sector is 6cm

bascially this with circle inscribed in it

r secttor is 6cm

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[royal loom]

For the "does it contain an element of order 40?" Could I say that, since every permutation can be written as a product of disjoint cycles, if an element has order 40 then it ca be written as a product of disjoint cycles, whose product of orders must be 40

and then since 40's divisors are 1,2,4, 5, 10 , 8, 10, 20, 40
the only way that this could happen is also if the total length of the disjoint cycles is 10 (since we're in S10) so the options are either

• a 10-cycle

• a 2 and 8 (disjoint) cycles

• a 1, 4, and 5 (disjoint) cycles

and nothing else is possible

and in all 3 cases, the 10-cycle will have order 10, the 2&8 will have order = lcm(2,8)=8 and the last will have order lcm(1,4,5)=20

none of which are 40

Would that be a complete proof?

@muted bear

[muted bear]

why me

[royal loom]

I just realized actually we can have multiple cycles of the same length still disjoint so this isn’t complete but maybe if I include them then it works

[royal loom]

Had to ask a sir

[muted bear]

i dont know group theory bruh you know this

[royal loom]

I was not aerate

[muted bear]

aerate?

[royal loom]

Consider this your introduction

[muted bear]

tell me what s10 is then

[royal loom]

Permutations of 10 elements

[muted bear]

oh

i think you should clarify that this means all partitions dont work

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[lapis cedar]

I can solve the p^2+q^2+r^2 and the pqr but for the others i'm stuck on how to approach them

[cedar kiln]

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[lapis cedar]

<@&286206848099549185>

[serene stag]

Mi hai chiamato?

[lapis cedar]

huh

english brother

<@&286206848099549185> anyone?

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[bright moss]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[crimson sedge]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

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[plucky holly]

I somehow got 4 and 8

4^3 = 64

[elder tapir]

If you had a number that ends with ba for example for n=4589 ba=89 (b=8 and a=9)

When cubing a number n, do we need to look at anything else than ba ?

[plucky holly]

ba?

i haven't heard of that term

[elder tapir]

It’s not a term

It’s just using letter instead of number

each letter represent a number

[elder tapir]

As in my example

Just the last two digits

for 678 ba=78, for 20300 ba=00

Do you get it ?

[elder tapir]

.

[cedar kiln]

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[sour heron]

friend me if ur very good at math

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[pseudo trellis]

Hey, I'm struggling to understand subsitution. I was wondering if someone would walk me through a few problems and just guide to see if I can start understanding it better.

[pseudo trellis]

This is the first problem

[wraith dagger]

Someone

[drifting marlin]

what about it is giving you issue?

[pseudo trellis]

I kinda understand what to set u to, but not really sure what to do after that point

[wraith dagger]

Someone

[drifting marlin]

well in general, once you've determined u, what do you do?

[pseudo trellis]

i guess that?

[wraith dagger]

Someone

[drifting marlin]

you've forgotten the dx twice, which is actually quite important particularly for u substitution

[pseudo trellis]

right

oooh

it's coming to me

[wraith dagger]

Someone

[pseudo trellis]

from here, i'm not quite sure what to do

would i just plug in the u value back in?

[dreamy void]

lol

[drifting marlin]

careful

those bounds are for x values

[pseudo trellis]

eeee

[wraith dagger]

Someone

[drifting marlin]

to get one of the bounds, yes

[pseudo trellis]

got it

[wraith dagger]

Someone

[pseudo trellis]

just put u= so I don't get confused in the future if i look back at this

[drifting marlin]

alright

[pseudo trellis]

from this point, would I split u^2/3 into two fractions for simplicity?

[drifting marlin]

if you want

[pseudo trellis]

then plug u back in?

[drifting marlin]

at this point you're left with something which you should be able to integrate

[pseudo trellis]

oh yeah

[wraith dagger]

Someone

[pseudo trellis]

what's the reason behind needing to change the -1 and 2 bounds?

i can't remember that

would it not work without changing them?

[drifting marlin]

because those original bounds tell you the x interval you're integrating over

but you're no longer integrating over x, you're integrating over u

[pseudo trellis]

i don't think that can possibly be correct

that's a large number

[drifting marlin]

again your bounds are from u = 1 to u = 10, not x

[pseudo trellis]

uhhh

so we don't sub x^3+2 back into u?

[drifting marlin]

if you changed your bounds, no

[pseudo trellis]

i see

[drifting marlin]

you could have written $\int_{x=-1}^{x=2} \frac{u^2}{3} \dd u$ and after antidifferentiating you would have to sub back

[wraith dagger]

Steakanator

[drifting marlin]

but since you changed the bounds, you don't

[pseudo trellis]

I got 111 as the answer

I see

[drifting marlin]

,w integrate x^2(x^3+2)^2 from -1 to 2

[wraith dagger]

Results provided by WolframAlpha

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

Requested by steakanator

[drifting marlin]

there you go

[pseudo trellis]

so i guess you don't need to make the bounds in terms of u, it just probably makes it more simple generally

[drifting marlin]

both roads lead to Rome

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[pseudo trellis]

Thank you!

[cedar kiln]

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[urban ermine]

This is a really big ask, but I figured I might as well. But is there anything else I should add to this sheet? It's basically a memory aid for an exam that I'll be taking in statistics, it's a year 1 course and I'll take any suggestions and lyk if it's something we've learnt/something I feel isn't needed as I know it well enough + don't want it to take up space

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@urban ermine Has your question been resolved?

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@urban ermine Has your question been resolved?

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[vapid gyro]

Guys how do I do this

Someone

[plain ridge]

a function has an inverse if it is bijective

which means f(x) has a unique value for every x

for example g doesn't have an inverse because g(1) = 1 as well as g(2) = 1

[urban ermine]

This is a really big ask, but I figured I might as well. But is there anything else I should add to this sheet? It's basically a memory aid for an exam that I'll be taking in statistics, it's a year 1 course and I'll take any suggestions and lyk if it's something we've learnt/something I feel isn't needed as I know it well enough + don't want it to take up space

Idk what I should add for what I said in the sis

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[outer heart]

how do i turn this into turning point form?

[outer heart]

the f is something else

not part of the question

[unborn verge]

do you mean like turning-point form of quadratic equations?

[outer heart]

im supposed to find the turning point of this

[unborn verge]

oh

[crimson sedge]

critical points?

[unborn verge]

if so, you should first differentiate V

and equate the expression on the other side to 0

solve the quadratic equation and those values of x are the turning points

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[crimson sedge]

explain the highlighted text😭😭

[crimson sedge]

can someone

highlighted text in the pee color

[static fern]

in other words, the outputs of a function depend on the inputs

[crimson sedge]

f(x) = x^2

[crimson sedge]

ik but can u tice an exampleee from the pic

[static fern]

tice?

[crimson sedge]

give*

[static fern]

the function in the pic appears to be y=x-10

so if we form a set of some outputs, it would depend on what we input into that function

[crimson sedge]

okay whats the difference bwtween the range and the domain

[static fern]

domain is the set of inputs and the range is the set of outputs

[crimson sedge]

okay

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[daring skiff]

Find x is the question

Pretty sure im missing something super obvious

Couldnt find that obvious thing after trying for five mins tho

[crimson sedge]

apply pytagoras in the 8 12 triangle

[daring skiff]

Ah thats not a 90 degrees triangle

Those 2 degrees with dots on them are the same

[barren hemlock]

just use similarity

[daring skiff]

How do i do that here

[barren hemlock]

well it would first be useful do label the points

i assume you know similarity?

just making sure

[daring skiff]

Yeh i do but there isnt enough

[barren hemlock]

ade is similar to abc

[daring skiff]

Given value here to do similarity

Oh from the top

[barren hemlock]

yeah

[daring skiff]

Didnt see that a was the same

[barren hemlock]

well should be easy after you see that

[daring skiff]

Found it ty

.close

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[half niche]

can someone help with this intergration

[stoic gale]

maybe start by factoring the top and bottom

[half niche]

i tried but its not working

[stoic gale]

what did you get

[void glen]

if it doesn’t work, just jump into long division then

[half niche]

it cannot be done like that but partial fraction

[void glen]

can’t do partial fractions if the degree of the numerator is greater than the degree of the denominator

it has to be less than the degree of the denominator

or else it won’t work

[half niche]

ohhh i didnt do that

[void glen]

yeah so i would suggest using long division first

and then you can prob do partial fractions

[half niche]

thanks

[void glen]

yw!

[half niche]

will the reminder or the factor be the numerator

[void glen]

remainder

and it’ll be divided by the factor

[half niche]

can you show me

[void glen]

like if the remainder is q and p(x) is what we are dividing by, then the term would be $$\frac{q}{p(x)}$$

[wraith dagger]

y0shi

[half niche]

okay

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[shy river]

How do I do this?

My teacher out and had a sub but that sub has a sub and didn’t teach us anything g

Can anyone help?

[surreal cave]

a tangent makes a right angle is a radius/diameter

[shy river]

It’s a right triangle

How does that help solve it tho

[junior dome]

can you tell me what is the length of that segment?

[shy river]

Idk what it is

If I did I could do a^2 + b^2

[junior dome]

well it is radius

[shy river]

So it’s 8?

[junior dome]

yeah

now just use pythagoras' theorem

[shy river]

Ty

Preciate you

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[crimson prairie]

Is there a field of math that has the ability to understand patterns in large amounts of data, like statistics, but doesn't lose any precision?

[main needle]

for like a general scatterplot of points? I don't think so. if they're in a clear pattern I guess it's functional analysis

[crimson prairie]

Can I give you an example of a problem that I'm facing

[main needle]

sure

[crimson prairie]

So say I have a huge physical system and The motion of every part of that system is determined

I know the general trend of that system but I don't know how it acts at every point

Is there a way of talking about it mathematically where I don't need to calculate every step of the way

Any ideas?

[main needle]

sounds like dynamics then

kind of a vague term lol

like #dynamical-systems

there's also control theory for robotics etc if that's what it's about

[crimson prairie]

No, I think Dynamics makes more sense

I'll look into it, thank you

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[crimson sedge]

please help

[cedar kiln]

@crimson sedge Has your question been resolved?

[carmine frost]

try splitting it up into two separate sums, the sum of the first n terms of 2^n (geometric) and the sum of the first nth terms of 3n (arithmetic)

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[keen tusk]

given the exact solution of this sum, how can i use it to solve for this other sum? i get how to do this if its simple like the original sum * 2 but when its more complex im not sure how to start

[solid juniper]

so basically it's asking you to evaluate the sum of the odd terms, yes?

[flint plinth]

probably you could find the sum of the even terms

and then...

[solid juniper]

why you gotta steal my thunder

[flint plinth]

wait who is this mysterious person

[keen tusk]

this would be the even terms right

[solid juniper]

yea

[keen tusk]

oh gotcha and then just do the exponent and then its just 1/16 times the original one

[solid juniper]

weird description but yea there should be a 1/16 somewhere

just to be clear you can write $$\sum_{n=1}^\infty \frac{1}{n^4} = \sum_{n=1}^\infty \frac{1}{(2n)^4} + \sum_{n=0}^\infty \frac{1}{(2n+1)^4}$$

[wraith dagger]

slayla

[solid juniper]

then compute the two of those sums you can compute

and then you'll know the value of $$\sum_{n=0}^\infty \frac{1}{(2n+1)^4}$$

[wraith dagger]

slayla

[keen tusk]

gotcha thank you. its always a simple solution just hard to see lol

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[crystal pendant]

hi im having trouble setting up this equation

[flint plinth]

do you have a picture of this A?

[crystal pendant]

Does this make sense

or should i simply plug my parameterized equation into the y and x component

and then just integrate?

[flint plinth]

yeah the picture looks ok

[crystal pendant]

it makes sense?

does the range work?

something someone said was like you should do -pi to -pi/2

[crystal pendant]

and then negate the integral

[flint plinth]

how did cos t + 1 become -cos t - 1

[crystal pendant]

i thought because the original integral has -x

[flint plinth]

i see, don't you need to swap -cos t - 1 with sin t?

the -cos t - 1 goes with dy, not dx

[crystal pendant]

but it turns into dt no?

does it matter

[flint plinth]

well the dot product won't be the same if you swap the components of the first vector

[crystal pendant]

and i would swap it because the y goes with dx?

[flint plinth]

yea

[crystal pendant]

so it should be (sint, -cost-1)

[flint plinth]

yes

[crystal pendant]

and b should be pi

and a should just be 3pi/2

or should that be swapped

[flint plinth]

you want to start at 3pi/2 and end at pi

so either integrate from 3pi/2 to pi, or do it from pi to 3pi/2 but change the sign of the integral

[crystal pendant]

Does this make sense

[cedar kiln]

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[limpid brook]

pls SOMEONE HELP

I HAVE A TEST TMR

PLS

ITS AN EASY QUESTION

BUT I DONT GET IT

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[cedar kiln]

Please don't occupy multiple help channels.

[sick ledge]

Could I show someone my work

This isnt right at all bit it is my attempt

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[rocky sigil]

Can you explain how to set this question up

[weak otter]

4x+y <= 1760

[rocky sigil]

Okay now how would you solve it after step by step please

[weak otter]

do you understand why this is

[deft saddle]

there is another equation that will make you find exact value of x and y

but learn how to set up question by coak

[rocky sigil]

This question pertains to utilizing quadratic formula or completing the square or solving by factoring how would I do it using any of those methods?

[weak otter]

first you need to understand why what i said is the case

[hazy dragon]

is this not a calculus question?

[rocky sigil]

No it’s intermediate algebra

[weak otter]

no because its easy to find the maximum of a quadratic

[hazy dragon]

i swear i got questions like this for my caluculus class where i needed to maximize functions using derivatives

[weak otter]

yes because in general its harder to find maximums

[rocky sigil]

Why did you set it up like that

[weak otter]

you only have 1760 feet of fencing

and according to the diagram, you need to use fencing on 4x and y

[rocky sigil]

Okay I understand now from there how would I move on? To figure out how to get 220 for my x value and 880 for my y value?

[weak otter]

you need to find the enclosed area in terms of x and y

by looking at the diagram

[hazy dragon]

@rocky sigil what class is this

[weak otter]

and then you maximize it

[rocky sigil]

Math 1010

1050*

[hazy dragon]

ok thanks

[rocky sigil]

Okay but like how would I do it though I understand how you set the equation up what I don’t understand is how to solve for x and y

Can you show me?

[livid hound]

what's your equation for the area

[rocky sigil]

4x+y=1760

Can you please show me how to do this question step by step?

[cedar kiln]

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@rocky sigil Has your question been resolved?

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[oblique frigate]

this is part of a wider solution in one of my calc 1 problems. i follow this right up until the third line. after that, i have no idea what's going on with the msup thing in red. and how the 8^2 got to the other side. i'm guessing it's some kind of cross multiplication, can someone confirm please

[oblique frigate]

and i'm guessing the h(t) was multiplied with the one that came before it to give h(t)^3

but then what happened to the remaining 3^2/8^2

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[royal loom]

Hi, I was hoping someone could check my proof and let me know if there is a better way of justifying why these two permutations are equivalent. I think my explanation was wordy but I wasn't sure how else to go abou it

[cedar kiln]

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@royal loom Has your question been resolved?

[royal loom]

<@&286206848099549185>

[spare spear]

?

[royal loom]

You assigned yourself the helpers role in id:customize, that is why you were pung when I pinged the helpers.

.close

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[gaunt bough]

what’s the antidiff of x^-1

[gaunt bough]

by logic would it not be -1

[mint garnet]

lnx

[gaunt bough]

Ohhh

alright

[upper abyss]

Try applying the power rule!

[gaunt bough]

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[jaunty mantle]

I think my method falls apart by dot point 3

[jaunty mantle]

Is there a better way to show equivalence classes?

Because if im not wrong, having sets with 2 elements doesnt increase the magnitude?

[candid marsh]

?

Sorry, not sure what you mean

[jaunty mantle]

|{1,3} U {2}| for example has magnitude of 2? or am i mistaken?

[candid marsh]

Cardinality of 3

[jaunty mantle]

Oh ok so my process works then?

[candid marsh]

Yeah

[jaunty mantle]

Ok thanks for clearing it up

[candid marsh]

It's just tedious
GL!

[jaunty mantle]

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[gilded remnant]

how do we know A/L = 1

[gilded remnant]

this is the rest of the question btw

and it was found limh->0 A/PQ = 1

but not sure how that relates to A/L

[cedar kiln]

@gilded remnant Has your question been resolved?

[gilded remnant]

<@&286206848099549185>

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[heady sierra]

you have to calculate the area under the curve

i dont know how to calculate the arc length of a parabola but its Area under the curve (between a and a + h) / length of the parabola section

is it defined as a parabola?

[cedar kiln]

@gilded remnant Has your question been resolved?

[gilded remnant]

its coshx so yes

[gilded remnant]

well yeah ik that already but the problem is the arc length

[heady sierra]

i think im learning that next lol

[gilded remnant]

might have smth to do with the previous part with the limit but then it would be a tangent not the arc length

[heady sierra]

what yr r u in?

[gilded remnant]

12

final year

<@&286206848099549185>

[candid marsh]

If PQ approaches A at infinatismally small values of h, then arc length is equivalent to the area under the curve

A/L=1

[gilded remnant]

woah really

is there an intuitive proof somewhere i can look at

maybe its kinda like the integration explanation by 3b1b but not completely understanding

[candid marsh]

It's more just
How calculus works?

[gilded remnant]

ok i think i get it now

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[crimson sedge]

help using u substitution

[crimson sedge]

am I doing it right?

help please and thank you

[coral jewel]

its better for you to usub the whole sqrt instead of just 2x-1

[crimson sedge]

okay wait

should i remove the square root in the denominator in getting the u?

or ill keep it raised to 1/2?

is this right?

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[spare spear]

dont say bulshit
i have the helper role

[junior dome]

?

[junior dome]

do you still need help?

[royal loom]

What?

[spare spear]

no

[junior dome]

didnt ask you

[spare spear]

هیچی کص ننت

😂😂

[crimson sedge]

yes

[junior dome]

.reopen

[cedar kiln]

✅

[spare spear]

the befor text you replay me

[junior dome]

are you allowed to use integration by parts?

[crimson sedge]

no

just u-sub

[junior dome]

ah

it would be best to split the integral first

[crimson sedge]

okay wait

like this?

[junior dome]

no

you cant do that

thats illegal

[crimson sedge]

ok sorry haha

[wraith dagger]

Dyssrupt

[junior dome]

i just separated the 2x^2 and 3

you can handle the second integral i think

[crimson sedge]

yes

[junior dome]

for the first, try subbing 2x - 1 = u^2

to get rid of the sqrt

[crimson sedge]

oh okay,

the procedure is still the same with this problem right?

@junior dome

[junior dome]

are you done with the previous one?

[crimson sedge]

no, just asking

[junior dome]

and yeah, this one would be the same but much easier i think

[crimson sedge]

ok okay, I still need to digest the procedure haha

thanks

[junior dome]

no prob

[cedar kiln]

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[signal current]

One way you could solve this is by finding out the distance of BC using pythag, then find out the angle ACB using trigonometry, then find the missing angle BAC. After that, you could construct the smaller triangle ABX and calculate the distance of BX using trigonometry

[river silo]

how?

well, i calculated up to angle acb

i got bc as sqrt 39

and angle acb as 38.682

[signal current]

still solve it, and it if looks like it might bisector you can use more decimal places

because this is a question asking whether x is the midpoint of BC you dont need complete accuracy

[river silo]

without knowing the angle axb

[signal current]

i dont think you need to know the angle AXB

because you just need to check if the distance BX = BC/2

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[signal current]

exponent both sides to e?

so that x is equal to e to the power of 15

[silver oxide]

ln(x)=15 is equivalent to $e^{ln(x)}=e^{15}$

damn forgot how use

oh wait

[wraith dagger]

Galaxy

[silver oxide]

there we go

[terse sigil]

nice

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[gray sorrel]

its a geometric series

and r is common ratio

s10 is sum of first ten terms

[lyric narwhal]

$S_{10}=244S_5$?

[wraith dagger]

kheerii

[gray sorrel]

yes

[lyric narwhal]

Okay

Use $S_n=\frac{a(r^n-1)}{r-1}$

[wraith dagger]

kheerii

[gray sorrel]

why not the other way round

in which it is 1-r^n

[lyric narwhal]

They’re both equivalent

You just multiply the numerator and denominator by -1

[gray sorrel]

$S_n=\frac{a( 1-r^n)}{1-r}$

[wraith dagger]

Durza

[gray sorrel]

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[tawny drum]

Differentiate xLn(x) - x

[tawny drum]

So i got xLn(x) = x*1/x = 1

-x = 1

so 1-1 = 0

but the answr was Ln(x)

[void glen]

so we have a product of two functions here

we need to apply the product rule

[tawny drum]

oh yeah oops

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[turbid mesa]

Help

[turbid mesa]

The hyperbolic distance between two points z1 and z2 is defined as tanh^-1(|z1-z2/1-z1conj×z2|)

Now is this the same as the length of the arc containing z1 and z2 of the euclidean circle they are part of in the unit disc?

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[crimson sedge]

the number of coins form a pattern. Draw an expression for the number of coins in the nth figure

[crimson sedge]

How do i do that?

[digital cliff]

1, 3, 5, ...

[neon void]

find a pattern first and try to explain it

[crimson sedge]

okok ty

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[sand cradle]

#1195800467880558672 message

[cedar kiln]

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[sand cradle]

<@&286206848099549185>

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[small igloo]

Odd question but I was calculating asymptotes and I want to know why this limit is +infinity and not -infinity

[small igloo]

can anyone try to explain me why

[deep void]

Look at what happens when we get closer and closer to 1 from both sides

[livid hound]

consider factorising the denominator

[small igloo]

I wanted a more algebric explanation

because I need to calculate the limit to make the graphic first

[crimson sedge]

do you know the wave curve method?

to determine the sign of a quadratic question ?

[small igloo]

No I wasn't taught that

[small igloo]

(x-1)(x-4)

@livid hound Hmm I think with this approach it makes more sense

[livid hound]

the wave curve method is essentially doing a rough sketch of a factored polynomial like above, focusing on x-intercepts and concavity

and it'll be pretty obvious where its positive or negative

for something purely algebraic,
you'd do sign analysis / make a sign table

[small igloo]

i think by factoring it makes more sense why the limit approaches +infinity to me

because if you substitue 1 approaching from the left both parenthesis will be negative so when you multiply it becomes positive

[livid hound]

wave curve method involves the first step of factoring too

its just what follows differs

[small igloo]

ok. i will try that as well

thanks for the help

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[small igloo]

i see

[livid hound]

something as rough / cursed as that is sufficient

(i deliberately made it look bad)

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[stiff surge]

bro

i heard

the first part = 0

and second part = 0

but why

[digital cliff]

if the product of two things is 0, at least one must be 0

[stiff surge]

OH

wait

ok

[stiff surge]

uhhh

i got 4

as one of the answers

[digital cliff]

doesnt sound right, how?

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[worldly chasm]

@crimson sedge to clear a complex number in the denominator you multiply by the conjugate

It's similar to clearing radicals in that way

(because that's what you're doing, sqrt(-1) and everything)

Yw

The complex conjugate of z = a + bi is z* = a - bi

So you would multiply by z*/z*

@crimson sedge

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[weak otter]

Yea

[worldly chasm]

When you simplify a fraction with a radical in the denominator, let's say a + b√2, you want to take advantage of the fact that (x + y)(x - y) = x^2 - y^2 so that you don't get the cross term.

So (a + b√2)(a - b√2) = a^2 - 2b^2

You multiply the fraction by (a - b√2)/(a - b√2)

Notice that I've said nothing about the numerator at all

Because we don't care about what happens to the numerator when we are rationalizing the denominator.

Because i = √(-1) it works the exact same way

So for your problem, which was to simplify (a - bi)/(a + bi) we are only concerned about the (a + bi)

No

Well, ok, yes it is the conjugate, but it is not the conjugate that we care about

We care about the conjugate of the denominator

Yes

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[analog sorrel]

What do I do from here? Do I add 2y on both sides?

[runic garnet]

when u mulitply over the dx, it becomes (5-2y)(dx)

see if u can go from there

[weak otter]

Is it chill to integrate both sides?

[gaunt radish]

Separate the equation so that u have y on one side and x on the other

[gaunt radish]

not untill u have x and y on one side

sorry on each side

[analog sorrel]

$\frac{2}{5-2y}dy = dx$?

[wraith dagger]

morphine_addiction

[gaunt radish]

so x on one and y on the other

[cerulean sail]

[the most likely way they want you to do it!]

[analog sorrel]

and then integrate to get -ln|5-2x|+c, sub in the 1 & 0 to get c

[runic garnet]

u sure it should be a x inside the ln?

check ur variable of integration

[analog sorrel]

ye i mean the other way around, i was doig it in my head

i got $\frac{5-e^{-x}}{2}-2 = y$

[wraith dagger]

morphine_addiction

[analog sorrel]

this is what i did but i got the wrong answer

[runic garnet]

ok so

-ln|5-2y| = x + C

and we have (x,y) = (0,1)

so

-ln|5-2| = 0 + C

-ln3 = C

thats how u should be finding C

[analog sorrel]

i subbed it in after rearanging

[runic garnet]

i dont think thats the right way to do it

[analog sorrel]

oh ye when i raised x to the e i didnt do it for c, so is $\frac{5-e^{-x}}{2}+ln|3|=y$?

[wraith dagger]

morphine_addiction

[runic garnet]

-ln|5-2y| = x - ln(3)

ln|5-2y| = ln(3) - x

e^(ln3 - x) = 5-2y

5- e^(ln3 - x) = 2y

y= 0.5(5- e^(ln3 - x))

y = 0.5(5 - 3e^(-x))

y = 2.5 - 1.5e^(-x)

that should be the answer

[runic garnet]

yea same thing

they just didnt distribute the 1/2

[runic garnet]

right here

[analog sorrel]

thanks

[runic garnet]

im not too sure on this

doesnt come out to the same answer

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[velvet egret]

Can someone explain how this big-oh works:

[velvet egret]

I don't understand how the min gets in there

[remote zinc]

What’s m

[velvet egret]

m, n are varaibles

I'm just having trouble with the second paragraph

@remote zinc Here's the full alg:

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[analog sorrel]

guys which one is is😭😭😭

[ancient lodge]

Send your solution and we’ll check it ig

[analog sorrel]

what even is the u witha dot

[ancient lodge]

the derivative of u

[analog sorrel]

so d/du?

[ancient lodge]

du/d(whatever u is a function of)

Normally this notation is used in particle motion

So take that as you will

[granite ether]

yeah its with like dx/dt for acceleration and stuff its weird using it here i think

not that exactly two separate examples

[analog sorrel]

i got e^k = u

[granite ether]

howd you get that am i being dumb

[dawn junco]

well do you want to say you have a correct solution or not ?
maybe you can just say yes and it will work out

[analog sorrel]

incorrect 😔

[analog sorrel]

i intergrated 1/u du = int k d then i got lnU = k and then i did e^k = u

[cerulean sail]

What even is this question

[analog sorrel]

did i atleast do the thing it was asking right or no?

[cerulean sail]

[though it's safe to assume that the dot implies the variable is t, so make sure you include that in your working out]

[analog sorrel]

new edexcel mcq 😎

how would i do this 1 marker? sub in t=0?

@ancient lodge how would i calc v as t aproaches infinite

[analog sorrel]

ye but like i cant sub in infinity in t

[ancient lodge]

That’s the point of taking the limit

$\lim_{t \to \infty} \frac{6t-3}{t+1}=6$

[wraith dagger]

Civil Service Pigeon

[analog sorrel]

how do u take the limit

[next nacelle]

Dont u just take yhe derivative of the problem?

Thats what i would do

No master here tho

[ancient lodge]

As in L’hôpitals?

That doesn’t rlly make sense here

[next nacelle]

Yeah alr, wasnt sure about it either

[ancient lodge]

Very hand wavy way: -3 and 1 are negligible as t approaches infinity, so the function can be approximated by (6t)/t = 6 for sufficiently large t

[analog sorrel]

i swear u can ignore the -3 and +1 as it dosnt make a difference as it approaches infinite and then y have 5t=v or am i making somethin gup

ye

[ancient lodge]

Slightly less hand wavy: $$\frac{6-\frac{3}{t}}{1+\frac{1}{t}} \to 6$$ since $\frac{1}{t} \to 0$

[wraith dagger]

Civil Service Pigeon

[analog sorrel]

ill take the more hand wavy way

but how do u go from 5t=v to v=6?

[ancient lodge]

What

[analog sorrel]

like u said the eq is bcically (6t)/t = 6 but where did u get 6 from

[ancient lodge]

Because 3(2t-1)=6t-3

[analog sorrel]

this one?

nvm im actually dumb

🤦‍♂️

a month away from alevels btw

ima ffali so bad

omd

clearing here i go

[ancient lodge]

https://media.discordapp.net/attachments/282239317966061568/1231360670835151011/emoji.png?ex=6636acfd&is=662437fd&hm=3fc80c55dc83ce3d82f7c34c6dbb14b0bd3cd775fed9d4dee700614aab72aa78&

[analog sorrel]

ight thanks for the help, u might see me here again like i have 12 more chapters to cover

[ancient lodge]

.close

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[split canopy]

.close

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[split canopy]

am i supposed to use trig identities?

i dont know what to do

[cerulean sail]

[may be worth saying what the shaded areas work out to be, in terms of theta and r?]

.reopen

[cedar kiln]

✅

[split canopy]

wait

180 - theta

and you can do the area formula

absin c

so

[cerulean sail]

half that but yep

[split canopy]

true

hm

[cerulean sail]

Also the other shaded region, what do you find that to be?

[split canopy]

a semi circle

[cerulean sail]

[remember that the angles are in radians]

[split canopy]

so pi - theta

[cerulean sail]

Well, not exactly semicircle-

[split canopy]

ture

true

[cerulean sail]

It's almost like a sector of a circle but with some part removed

[split canopy]

thats right

so area of a sector - area of unshaded triangle

ok

now i have the two equations

i set them equal to eachother and solve?

alright

give me a second

[cerulean sail]

[also remember you'll need an identity!]

[split canopy]

hm

i got to 0 = theta

so yeah

ill re-try

i think the mistake is done when i get to sin (pi - theta) = theta - sin theta

i distribuite sin and cancel out the two sin theta

then its sin pi = theta

0 = theta

[cerulean sail]

Wait how do you mean distribute sin? As in you do "sin(pi - theta) = sin(pi) - sin(theta)"?

[split canopy]

oh shit

hahaha

you cant do that

yes thats what i meant

but if not

then what

[cerulean sail]

You can't Buutttttt sin(pi - theta) simplifies nicely to something...!

If I write it in degrees, sin(180 - theta), would you know what that becomes?

[split canopy]

still no

honestly

any postive number

since theta is larger than 0 and less than pi

[cerulean sail]

awwww

Worth checking up on identities, but you have sin(pi - theta) = sin(theta)

There's a whole list of them around

[split canopy]

yes

ill have a formula booklet for this test

but this one doesnt show up

maybe its just assumed that i should know this

anyways

sin theta = theta - sin theta

hence 2 sin theta = theta

thank you for your time

and patience🙏

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[scarlet flame]

how do I do this with x
Exponential Decay (increasing form)

[scarlet flame]

without a cap

[hollow totem]

Approximation?

[scarlet flame]

wdym

oh

yeah

[neon void]

so a logarithm?

[scarlet flame]

yeah that

lemme see if i can make a formula for it rq

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[lost delta]

do all log functions have their range equal to R

[dreamy void]

you mean log(x) with any base?

then yes

[lost delta]

yes

natural logs?

any exception I need to know

or just to know

looking at a graph

how do I know if the range is all R

[humble karma]

If you're familiar with the change of base formula, then you can use that to write any logarithmic function as a scaled version of ln(x). Since ln(x) ranges through R, then scaling it doesn't matter, and so any log function will cover all of R.

[lost delta]

just as a rule

[tame knoll]

if the basic log function has a restricted domain it will not have a range of R, otherwise its range is R

[lost delta]

all log functions have restricted domains

[tame knoll]

restricted domain on the positive side of x-axis

[lost delta]

yet all of them (as far as I am aware) have range = R

just as a rule

is it safe to say

that all log functions

have a range of R

[humble karma]

Yes.

[lost delta]

if not

what are the exceptions

[tame knoll]

i.e. consider log(x) for (0, 3], it doesnt have a range of R

[lost delta]

the same way that the domain for all exponential functions

is R

the inverse (logs)

have range R

okay I answered myself

[humble karma]

Yes well they are inverses of eachother, so the range of one is the domain of the other.

[lost delta]

allright thanks for the help all of you

oh one last thing

just curious

why are natural logarithms everywhere

why are they so important

[tame knoll]

im pretty sure they model alot of things in nature

[cedar kiln]

@lost delta Has your question been resolved?

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[lost delta]

.close

[swift saffron]

Suppose s(n) denote the sum digit of n. Prove that for every natural number x,y :
s(xy) <= s(x)s(y)

[neon void]

sum digit is a big hint

it's definitely gotta do with log10

$s(n) = \floor{\log_{10}{n}} + 1$

ok

[wraith dagger]

mochaccino

[plain ridge]

are you sure there's a closed formula?

[neon void]

it only needs to work for natural numbers

[neon void]

oh its sum digit not digit count

welp nvm then it's not logarithms

welp thought the bot would do smth with that

[swift saffron]

I just want to check my work

[neon void]

but there is a formal defintion for s(n)

[swift saffron]

This is my answer :

From s(xy) <= y s(x), implies s(yx) <= x s(y).
So,
s(xy)s(yx) <= y.s(x) x.s(y)
And
s(yx) <= xy,
s(xy) s(yx) <= xy s(x) s(y)
s(xy) <= s(x)s(y)

[neon void]

why do you assume the first statement

[swift saffron]

Its always correct by induction

[neon void]

ok

[swift saffron]

Is this allowed :
s(xy)s(yx) <= xy . s(x)s(y)
because i want to prove that s(xy) <= s(x)s(y). So s(yx) <= xy

<@&286206848099549185>

[queen heath]

why did you ping helpers

you have mochaccino

oh no my bad im stupid

assumed she was a helper

[cedar kiln]

@swift saffron Has your question been resolved?

[cedar kiln]

@swift saffron Has your question been resolved?

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[vocal patio]

when performing definite integration, with u substituion, do you still need to substitute the value of u back into the equation when removing the integral?

[vocal patio]

so here, the value of u was substituted back in

[royal loom]

If your original question asks for an answer in terms of x, then yes