help-33

[plain trellis]

kk do you know midpoint theorem?

[still temple]

no i dont

[plain trellis]

https://en.wikipedia.org/wiki/Midpoint_theorem_(triangle)

[still temple]

i havent learnt it in the higher gcse aqa maths spec

thabk you so much

[plain trellis]

apply the theorem

[still temple]

okay

how does this apply to squares though

[plain trellis]

you find the lenght of the side?

construct the diagonal of the first square and try to apply the thoerem to find CD

[still temple]

OH okay

[still temple]

but like what does “third side” mean here because im not starting off with a traingle theres just one being created

[plain trellis]

let the original(big square) be ABMN

construct BN

try to apply it to the triangle ABN

$CD = 1/2 BN$ right?

[elfin berry]

Annie Maqionde

[plain trellis]

contiune with the same steps

[still temple]

are m and b on the same line

or diaganol

[plain trellis]

BN is one diagonal

[still temple]

okay okay got it

[stoic saddle]

\frac{1}{2}

if youre gonna use LaTeX then use actual LaTeX

[wet flint]

its actually pretty easier then u think

hold up lemme write

srry im rusty my hand writing sucks 😭

[marsh citrus]

@karmic jungle Has your question been resolved?

[marsh citrus]

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[misty vapor]

if
abc <= 1/8
and
a+b+c >= 3/2
for positive reals a,b,c
would it be right to conclude
(1+abc)/(a+b+c) <= 3/4?

abc <= 1/8
1 + abc <= 9/8
(1+abc)/(a+b+c) <= 9/(8(a+b+c)) <= 9/8 * 2/3 = 3/4
it seems right to simply plug in the inequalities, but I just want to verify

also I got those two inequalities from (a+b)(b+c)(a+c) = 1

[misty vapor]

so
[(a+b) + (b+c) + (a+c)]/3 >= ((a+b)(b+c)(a+c))^(1/3)
thus a+b+c >= 3/2
and
1 = (a+b)(b+c)(a+c) >= 8abc
thus abc <= 1/8

[marsh citrus]

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@misty vapor Has your question been resolved?

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[inland grove]

i don't know what to do for the right side

[inland grove]

i could brute force it and just expand ig?

but im guessing theres something better I could do

I've simplified the left to -(9^10)/2 + (9^10)(sqrt(3))i/2

[devout mauve]

you can simplify sqrt(243)

and then you can convert into polar form

(or recognize the number)

[inland grove]

? right

i've simplified the right side

i just don't know what to do with the left

[devout mauve]

well ok read what you wrote

[inland grove]

uh
(z/2 - 1 + 2i)^4 = -(9^10)/2 + (9^10)(sqrt(3))i/2

altho i'm guessing I simplified it wrong?

[runic temple]

$(a+b)^c\neq a^c+b^c$

[elfin berry]

726=|4 step 3D lattice trails|

USS-Enterprise

[inland grove]

I was using this corollary of De Moivre's Theorem

unless its somehow not applicable here?

[runic temple]

oh you already did the demoivre part

mb

[main idol]

Like i said last time you should have taken the fourth root of both sides first

[inland grove]

I tried last night but tbh I'm not sure if what I was doing was right at all

and I just got really confused

[runic temple]

dont forget that there will be 4 solutions

[inland grove]

Im assuming I'm supposed to use this

but idk how to do it?

unless pheta is the same as the one found on the right hand side? but then that would be finding the root of the left hand expression not z?

[main idol]

,tex .exp rules

[elfin berry]

riemann

[main idol]

10 / 4 = 2.5

[inland grove]

Can I just take the 4th root of both sides?

I thought roots were only defined for reals

not complex numbers

[main idol]

Yes that's what this means

[main idol]

This is wrong

[inland grove]

oh ok

my textbook says this about root notation. Does this only applyfor the notation? but otherwise the function works exactly the same?

[main idol]

Wut

You use demoivre

[inland grove]

well yes but I think this passage was why I was confused?

I am not really good at fully understanding detials

ok so is root as an operation defined exactly as it is for real numbers?

[marsh citrus]

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[dire harness]

can anyone help me whith this 🙁 #math-discussion message

[waxen dust]

Just post the question here

Makes it easier for helpers

[dire harness]

I'm strugling to understand 5on the -1 th or on the -2nd

1 min and i will sent the pic

[waxen dust]

What is 5^3

[dire harness]

5times 5times5 (3 times 5 times 5 times5)

[waxen dust]

Okay

We agree, 125

Right

[dire harness]

the question is : how can I solve x^-n

[waxen dust]

I know

We are getting there

[dire harness]

okay

[waxen dust]

Okay

So 5^3 = 125

What is 5^2

[dire harness]

25

[waxen dust]

Okay

And how did we get from 5^3 to 5^2

Or from 125 to 25

[dire harness]

/5 I think

[waxen dust]

Yes, we divided by 5

125 / 5 = 25

What about 5^1

[dire harness]

5^1is 5

[waxen dust]

5^1 is 5

Again, we divide by 5

25 / 5 = 5

Now

What is 5^0

[dire harness]

1

[waxen dust]

Exactly

[dire harness]

5/5

[waxen dust]

Notice again

5 / 5 = 1

And then

What would 5^(-1) be

[dire harness]

so I just divide 1 by 5?

[waxen dust]

Yes, exactly

We keep dividing by 5 when we go lower and lower in the power

So 5^(-1) is just 1 divided by 5

or 1/5 or 0.2

And then 5^(-2) is that (1/5) divided by 5, or 1/25

[dire harness]

and then 5^-10 is 1 divided by 5 10 times?

[waxen dust]

$\frac{1}{5^2}$

[elfin berry]

USS-Enterprise

[waxen dust]

Yes

We notice

$5^{-x} = \frac{1}{5^x}$

[elfin berry]

USS-Enterprise

[waxen dust]

We had x = -1, which was 1 / 5^1 or 1/5

[dire harness]

ohhhhhh

[waxen dust]

Then for x = -2, we have 1/5 divided by 5, or 1/5^2

[dire harness]

my teacher had talked about something like
1

10

5 I may not be correct

[waxen dust]

This also works for any base $m \in \mbb{R} - {0}$

[elfin berry]

USS-Enterprise

[waxen dust]

Not just 5

[waxen dust]

I can't say I have heard of this before

So, in general

$m^{-x} = \frac{1}{m^x}$

[dire harness]

okay 🙂 thanks

[waxen dust]

$m \neq 0$

[elfin berry]

USS-Enterprise

USS-Enterprise

[waxen dust]

And as I showed, think of it as just keep on dividing by that number

There's no reason to stop after you enter negative numbers

So let me give you an example then

What is $7^{-3}$

[elfin berry]

USS-Enterprise

[dire harness]

soo 5^-1 ------>1
-----
5^1 which equals to 1/5?

[waxen dust]

Yes!

[dire harness]

1/7/7/7?

[waxen dust]

Yes exactly

Just usually we then write 1/7^3

But the idea you wrote is exactly the point - we keep dividing by 7 (three times)

So $7^{-3} = \frac{1}{7^3} = \frac{1}{343}$

[elfin berry]

USS-Enterprise

[waxen dust]

Also, important

This doesn't just work for natural powers like 3, 6, 7, 12

It works for all real powers

$5^{-2.5} = \frac{1}{5^{2.5}}$

[elfin berry]

USS-Enterprise

[waxen dust]

The idea is the same

But we can't really visualize dividing by 5 2 and a half times if you get me

[dire harness]

okay 🙂

[waxen dust]

🙂

Hope I've been of help

If that's all, please .close

[dire harness]

.close

[marsh citrus]

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[ember saffron]

Need help with Problem 19

[ember saffron]

Is this a related rates problem?

<@&286206848099549185>

[unique bough]

I don't think so. You could use MVT to indicate that there must be a period of time where the rate is greater than 225,000 gal/min

[ember saffron]

Ye but what is the function????

MVT is f(b)-f(a)/b-a

[unique bough]

MVT means there exists some $c$ in $(a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b-a}$

[elfin berry]

nhknyugn

[unique bough]

Here f' is the rate function.
Assume the initial volume is v0. You can calculate the volume after 24 hours using the given rate.

[ember saffron]

Ye but what is the function?????

[unique bough]

Function what

[ember saffron]

What is f(b)

and what is f(a)

[unique bough]

f(b) is the volume of water at time b

Maybe my naming is a bit confusing

[ember saffron]

I'm so lost

I thought this would be a related rates problem because it talks about the rate in excess of 225,000 gal/min

[vivid basin]

i think f'(c) would be the 22500 gal/min and you would have to find a value for lke f(b) thats larger?

[unique bough]

Now let V(t) be the volume function, then V'(t) be the rate.
You have to show that there exists a t0 in the interval (0, 24 hours) (a day) such that V'(t0) is greater than 225,000 gal/min

[vivid basin]

^^

[unique bough]

You can easily eliminate the initial volume in the numerator since it is just the change in volume.

Also don't forget to convert the unit

[ember saffron]

ok hold on

So

f'(c) = 225,000 gal/min right?

And 1440 acre - feet is f(a)?

Then what is the 24 hours doing there

oh wait

[unique bough]

How did you come up with this number...

[ember saffron]

thats just a

[ember saffron]

Its in the problem

Look at second image

[unique bough]

^

[ember saffron]

ye so isn't f'(c) = 225,000?

[unique bough]

I mean there must be some V'(t0) with t0 in (0, 24 hours) such that V'(t0) > 225,000 gal/min

[ember saffron]

I'm still lost

What I was gonna do was

225,000 gal/min = f(0) - 1400 / 0 - 24

Then solve for f(0)

[unique bough]

Use MTV to calculate $V'(t_0) = \frac{V(24\text{ hours}) - V(0)}{24 \text{ hours} - 0}$

[elfin berry]

nhknyugn

[unique bough]

We don't know what t0 is but we know such rate exists

[ember saffron]

Ye but what is the volume!

Whats the formula!

There is no shape given

[unique bough]

You have the rate and the time

1400 acre-ft in 24 hours

What's the volume then

[ember saffron]

1400/24?

so

58.3

[unique bough]

Oh wait

Acre-ft is already the volume...

You just need to convert it to gal

Then the denominator convert it to minute

[ember saffron]

so multiply by 60?

[unique bough]

Yeah

[ember saffron]

Ok I got 1440

So it goes from 0 to 1440 right?

So a = 0

b = 1440

[unique bough]

Yeah

[ember saffron]

What now?

It says I have to convert

so do I do 1440 x 7.48?

or 1400 x 7.84?

[unique bough]

Note that

1 acre-foot = 43.560 ft^3
1 ft^3 = 7.48 gal

[ember saffron]

Bro I'm so lost

[unique bough]

Just calm down it's just a simple conversion...

[ember saffron]

1400 arce -ft x 43.560?

also its not 43.560

its 43,560

[unique bough]

Yeah I get used to use comma for the decimal mark

Acre-ft * 43560 => ft^3
ft^3 * 748 => gal

[ember saffron]

I figured it out through Quizlet

But don't really understand

Why did they mulitply everything out?

[unique bough]

Don't you know how unit conversion works?

[ember saffron]

I do

Just confused why its all multiplied

[unique bough]

Because the question gives you a volume in acre-ft then asks for gal

[ember saffron]

I see

thank you

.close

[marsh citrus]

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[marsh citrus]

@ocean mist Has your question been resolved?

[halcyon spire]

do you still need help

[marsh citrus]

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[buoyant jetty]

how do i find theta

[elfin cairn]

What have you tried

[buoyant jetty]

soh cah toa

[pseudo flax]

do you know the unit circle definitions of sin, cos and tan?

[buoyant jetty]

i don't want that

[novel juniper]

then how do you expect to do this

[buoyant jetty]

how do people before unit circle existed did it?

[limpid dust]

When trig was created it was a fact

What is that is it sin(theta)=-1

[buoyant jetty]

.?

[upbeat python]

The unit circle is a direct result of trig/it’s almost a part of its definition

[pseudo flax]

just think about the fact that when we restrict ourselves to right triangles, it simply isnt possible to produce negative ratios by taking any two sides

[limpid dust]

Yeah exactly

[pseudo flax]

so how do you plan on figuring theta using a right triangle and soh cah toa

[buoyant jetty]

worst thing i can think of its finding the absolute minimum of sine

[upbeat python]

What do you mean by absolute minimum?

[buoyant jetty]

optimization, if you have seen single variable calculus

[upbeat python]

sin(x) is a periodic function

The absolute minimum of sin(x) in the sense of the smallest possible nonnegative value of sin is just 0

because $0 \leq| \sin x | \leq 1$

[elfin berry]

plantsyavi

[buoyant jetty]

why you put the absolute value

[pseudo flax]

okay if you want to consult the graphs then your solution set will be whenever the curve of sin theta and the line y=-1 intersect. but how do you know about the periodicity and stuff without the unit circle?

[upbeat python]

Because you said absolute minimum

Agreed :) basically I think me and @pseudo flax are both saying is the unit circle is super important because it gives a great visualization of the properties of trig functions

[buoyant jetty]

i mean the global mínima

[upbeat python]

The lowest possible value sin x can give is -1

And the highest is 1

[buoyant jetty]

yes at what x

[pseudo flax]

oh sorry mb. miscommunication @upbeat python i thought they wote that. again, im sorry. please carry on

[upbeat python]

$x=3\pi / 2$ gives -1

[elfin berry]

plantsyavi

[upbeat python]

And of course all periodic terms as well

So you can add 2pi*n for whole number n

And $x=\pi /2$ gives 1

[elfin berry]

plantsyavi

[upbeat python]

Again you can add a multiple of 2pi due to periodicity

The unit circle is great at visualizing this

I think you need to think of sin and cos as more general functions than just in soh/cah/toa

[buoyant jetty]

sure but without unit circle

[honest creek]

absolute minimum of sin(x) is -1; it’s not the “smallest possible nonnegative value of sin”

[upbeat python]

I interpreted it differently but you’re right yes :)

[upbeat python]

this equation still holds though

[buoyant jetty]

using a little bit of calculus you don't have to memorize the unit circle for this, though this is the first time soh cah toa has failed me

[upbeat python]

I’m not sure but I feel like you’re viewing the unit circle as a unique concept

But it’s not

The unit circle just gives a visualization for sin/cosine and the values they can take

[buoyant jetty]

unique? i just don't like memorizing stuff

[upbeat python]

Its a circle with radius 1 centered at the origin

There’s not much to memorize

Do you understand what the unit circle tells us

[buoyant jetty]

a bunch of values to memorize is what unit circle is afaik, you turn the x coordinate as cosine of theta and y coordinate as sin theta, each x,y gives the outputs of using some theta to both

@upbeat python

[upbeat python]

yup but the important thing is not the values you get but what the circle tells us

The fact that it’s a circle means that once you reach the angle 2pi the functions ‘restart’ (what I’ve called periodicity)

You can also see symmetries in Sinx/cosx

[buoyant jetty]

what is your point

[upbeat python]

Using the unit circle you can see in 2 seconds what point gives sin(x)=-1 or sin(x)=1 etc

No calculation needed

[buoyant jetty]

do i need to memorize something?

[upbeat python]

You can also understand why $\sin(x+2\pi) = \sin x$

[elfin berry]

plantsyavi

[upbeat python]

No

[loud summit]

you've memorized ssol. For cosx=0 here...

[buoyant jetty]

you can use soh cah toa for this

is not that i memorized, i applied soh cah toa in my head

[upbeat python]

Soh cah toa is so limited

It’s like a special case

[loud summit]

u dont need to memorize it tho, it js sits in ur head when you've used it a no. Of times

[upbeat python]

Like it’s just a circle

99% of the time you don’t need to know obscure angles

[loud summit]

also i hv no idea whts actually going on here lol

[buoyant jetty]

imo soh cah toa is useful for almost everything

i just don't like memorizing stuff

[loud summit]

me too icl

[buoyant jetty]

and i am way past trig class so i can't practice unit circle problems in order to memorize by using it

i just encounter some trig or another in my daily math

[buoyant jetty]

also i didn't liked using optimization for this

but oh well, memorizing the unit circle is quite a task by itself

[upbeat python]

Haven’t used soh cah toa since geometry

But when dealing with functions you have to look at the values of x instead of ratios

[loud summit]

MMaybe You could've used sin(pi/2)=1

multiply both sides by -1

[upbeat python]

A very simple counter example to this is how you would deal with sin of an irrational number

[buoyant jetty]

yes and sin(-x)=-sin(x)

[loud summit]

yep

altho 3pi/2 is a standard angle itself

[upbeat python]

this is also a property you see on the unit circle

[loud summit]

turu lol

[buoyant jetty]

like what for example?

and does unit circle cover that?

[upbeat python]

In a way yes

Practically though you wouldn’t look for the angle on the unit circle

Anyways I don’t think it’s worth arguing but the unit circle is a natural way to understand properties of trig functions

[buoyant jetty]

surely, i just don't know how to learn it without memorizing

[upbeat python]

It’s more about understanding than memorizing

[buoyant jetty]

how so

[upbeat python]

e.g. whenever the unit circle crosses the x axis, at that point sin(x)=0

so you can look for the values of x where that occurs

similarly for the y axis,

[loud summit]

and revising as well....

[upbeat python]

periodicity follows directly

understanding things like symmetry: why is sin(-x)=sin(x)? cos(-x)=cos(x)?

These are all visualized very nicely with the unit circle

So a simple diagram tells us a lot about different properties

[buoyant jetty]

how to derive the unit circle for any angle of choice

okok i think what i learnt today was very useful

you can derive any angle from the unit circle without memorizing

you just have to interpret the soh cah toa, use Pythagoras theorem

here the hypothenuse is 1 and the opposite is -1

i always thought what? here is where soh cah toa fails, because we can't have negative sides in triangles

but if you think it through using the unit circle, the adjacent is cosine and the opposite is sine

then it makes perfect sense now

and everything clicks together now @upbeat python @loud summit

Hopefully someone understands what i did here to find the alpha when sin = -1 and cosine is 0

i say cosine is 0 because that's what we got from Pythagoras

i just used the unit circle without memorizing anything

all you need to know is a parametrization of a circle of radius 1 with (x,y) = (cos(t), sin(t))

[loud summit]

i do understand xd

[buoyant jetty]

i appreciate it

today was really productive because for all my life i thought unit circle was useless

[loud summit]

+1109 knowledge acquired

[buoyant jetty]

people memorize the unit circle, but at the end of the day you just need to b understand how it works and you don't need to memorize anything

[loud summit]

ima go my head hurts ATB

[loud summit]

yeah well idk what memorization you're taking abt tbh.

[buoyant jetty]

back when i was at high-school i memorized it

[sacred idol]

now you can see sin(3pi/2) = -1 with your third eye

congrats

[buoyant jetty]

i think today was really productive

.solved

[marsh citrus]

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[red nimbus]

it's a whole unit of a circle

[marsh citrus]

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[hollow solstice]

can any1 help me with part a?

[gentle current]

use binomial formula

-p*3/2=-24

p=16

?

https://share.google/images/RRjVMBZtefnrfcfHW

@hollow solstice

[marsh citrus]

@hollow solstice Has your question been resolved?

[marsh citrus]

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[onyx rivet]

"the fish population in a lake is dying off at a continuous rate of 6%. How many remain afyer 5 years if the population started at 1300?" which formula is supposed to be used for this question?

[wet rock]

Are you familiar with the continuous exponential decay formula?

Because if so you just plug in everything and you are set

If not, it is

$$
P(t) = P_0 , e^{rt}
$$

[elfin berry]

zedias

[onyx rivet]

okay thank you

[wet rock]

np

[onyx rivet]

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[onyx rivet]

"Your bank currently has $1,250. You are getting 1.6% interest annually. How much will you have in 3 years?" does the word annually imply that it's compounded formula or is there not enough info to say compounded and it's just the simple interest formula?

[indigo nest]

There isnt enough info to do either tbh

I would do an annual compounding personally

But the question is ambiguous

[onyx rivet]

okay thank you

[onyx rivet]

I think the questions are ambiguous because they expect us to only do what we learned in the unit,

[indigo nest]

Maybe

[onyx rivet]

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[novel juniper]

Is this asking me to show there are exactly 2

[calm grove]

Let us imagine what a generator for such a group would look like. We need exactly one element A, with the property that A^3 = id (that element generates our subgroup). Now we can change our question to: what properties do the other generator elements need? What is not allowed for them?

[novel juniper]

hmm

[novel juniper]

Can't I just use the fundamental theorm of finite abelian groups

The issue ofcourse is that , doing so is VERY tedious

[calm grove]

Oh, I have no idea what this is. We didn't got there in our Algebra module.

[novel juniper]

mhm

[novel juniper]

Is this asking me to show there are exactly 2

[calm grove]

My idea would be that we have 27 * 2 * 2 and 27*4 as the only two ways to factorise 108 without having another 3. I don't know how to express that in the words of group theory, sorry.

[calm grove]

I think so, because there are only two, I think.

[novel juniper]

mhm

[novel juniper]

sure, but those have multiple subgroups of order 3 I think

[calm grove]

Z/27Z has only one subset of order 3, right? 9Z/27Z

[novel juniper]

oooh

quotienting is a smart way to do it

[regal coral]

if the 27 was broken up, there would be more than one subgroup of order 3

[novel juniper]

noted

[calm grove]

I have to leave, I hope I was able to help at least a little bit. Have a great day, you can get this.

[novel juniper]

thanks!

[novel juniper]

Thanks!

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[fast geyser]

can someone help me teach how to evaluate? I do not understand the explanation from my notes

[fast geyser]

number 2

pls ping thx

[upbeat python]

Instead of brute force, use the fact that you are given the eigenvalues and vectors of the matrix

[fast geyser]

well yes my eigenvalues are 1+i, 2, and 1-i

and matrix is on the left

[upbeat python]

Yep, what does that tell yoy about the matrix you’re looking for

[fast geyser]

Av = lambda x v

are we looking for v?

so A is the matrix on the left

[upbeat python]

We are looking for A

You already have the vectors v

I should clarify: one way you can solve it is to actually compute the matrix products

And calculate the inverse etc

But this method is sooo long

What I suggested could make it shorter but im not sure, worth a shot :^)

[upbeat python]

Using this, if you just set A to an arbitrary matrix (all you know is that it’s 3x3), you can solve some equations

[fast geyser]

wait i can show u how the note solves it

its less than halff of the page bruh

but brb

there is actually more stuff up there, let me know if u wanna look

[upbeat python]

Ooo neat

Also sorry I misunderstood the problem, thought we were looking for the original matrix

This looks like it’s for a different problem but you can apply the same principles

You separate the matrices (all 3) into sums of a real matrix and a complex, and that’s basically it!!

[fast geyser]

then how does the eigenvalue change

the middle one

and how about the last column of the left matrix? Why is it not separated and gone

@upbeat python this is hard, I prefer multivariable 😭

[marsh citrus]

@fast geyser Has your question been resolved?

[fast geyser]

<@&286206848099549185> help numero 2

pls ping me when answering thx

[blazing pulsar]

If your textbook doesn't carefully derive this construction from scratch, I would just follow the algorithm you are given.

You should have somewhere that for a real valued matrix, an eigenvalue a+bi comes in conjugate pairs and can form a jordan block where a11 a22 = a, a12 = -b and a21 = b like you see in that picture.

[marsh citrus]

@fast geyser Has your question been resolved?

[marsh citrus]

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[novel juniper]

I was thinking we first use 1st iso to determine the size of the kernel for a given size of image

$|Z_4/ Ker( \phi)|= |\phi(Z_4)|$

[elfin berry]

wai

[novel juniper]

I'll be back in a bit

Sorry

.close

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[novel juniper]

.reopen

[marsh citrus]

✅ Original question: #help-33 message

[novel juniper]

okay, i'm back

is my idea fine

[devout mauve]

you could just list them...

[novel juniper]

the homorphisms?

[devout mauve]

just knowing the (size of) kernel isnt enough to determine the homomorphism anyway

[novel juniper]

okay, so I just list them out

The trivial homomorphism. The mapping taking 1 to (1,0), the mapping taking 1 to (1,1) and the mapping taking 1 to (0,1)

this uniquely defines the homomorphism as Z_4 is cyclic so its image is determined by its generator

[devout mauve]

yes

[novel juniper]

this is identical I suppose

[devout mauve]

yes

[novel juniper]

thanks

.close

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[silver shuttle]

ned help

[silver shuttle]

VA s are -1 and 5

i think 😭

horizontal asymptote is -2

[fair pond]

What about the roots?

[silver shuttle]

x is (1,0) and y is (0,1)

[glossy flint]

Huh? Nope

[silver shuttle]

hm

both wrong?

mb

its kinda hard to see this stupid graph

[glossy flint]

By the way, roots mean intersections with the x-axis only

[silver shuttle]

hm

[glossy flint]

Roots or zeros, those are synonyms fyi

[silver shuttle]

(2,))

[glossy flint]

Nope

[silver shuttle]

wait

0 and 2

3*

0,0 and 0,3

i mean

the other way

but u get it

3 for x

[glossy flint]

Yup awesome 👍

[silver shuttle]

wait sooo

omy

tysm

helped me sm

i ned more help 😭

are the roots here (0,-1) and (2,0

sum1

any1

[marsh citrus]

@silver shuttle Has your question been resolved?

[main idol]

roots mean y = 0 so there's only one root

[marsh citrus]

@silver shuttle Has your question been resolved?

[silver shuttle]

ty

[silver shuttle]

steel no

if the root is (0,2)

im confusious

broo

[kind harness]

What are you confused about??

Suppose you got a function y = x + 2

What are the solution of this equation?

@silver shuttle Do you know the difference between solution and root of an equation?

[marsh citrus]

@silver shuttle Has your question been resolved?

[silver shuttle]

mkay

[slim surge]

find the roots

then you should be able to give a first guess for the factored form of the polynomial

[marsh citrus]

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[arctic steppe]

Im not sure if im doing this correctly but this is what I have so far

[cunning fiber]

$x=0$ is fine. Not sure where $x=\frac{1}{2}$ came from.

[elfin berry]

Civil Service Pigeon

[late geode]

$x^{1/2} \redneq x \cdot 1^{\frac12}$

[elfin berry]

ραμOmeganato5

[arctic steppe]

Because i changed square root of x

To x of one half but then i wasnt sure how to factor that

So i substituted 1 for x but im pretty sure thats wrong 😭

Im not sure what to do there

[late geode]

consider factoriing out x^(1/2)

[arctic steppe]

Thats 1/2x^-1/2 right

?

[late geode]

no

which part are you referring to

[arctic steppe]

The factor if x^1/2

[late geode]

$-x^2 + x^{\frac12} = x^{\frac12} \cdot \what$

[elfin berry]

ραμOmeganato5

[arctic steppe]

?

.close

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[sonic basin]

Anyone can tutor me only one 2 hour session will pay $50. Calculus Math 157 university

[sonic basin]

Just need help with some topics

[crystal lintel]

blud

[vague jay]

yo

can I do this tmrw

for you moby

i'm down

[marsh citrus]

@sonic basin Has your question been resolved?

[sonic basin]

sure

What time are you free

[marsh citrus]

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[spiral epoch]

e

[spiral epoch]

.claim

why do i need to solve it like this

ohhh nvm

i understand

the triangle is ADE

questions says AED and ABC are similar

not ADE

[stoic saddle]

not a command

[spiral epoch]

.close

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[lethal steeple]

how to solve this one

[distant saffron]

how to simplyfy this

[crude hearth]

is your goal to simplify the expression?

[lethal steeple]

yes

[main idol]

Use 98 = 2 * 7^2

[crude hearth]

oh, okay

try to factor the bottom and top expression

and factor using difference of squares

then, some terms will cancel out

@lethal steeple do you still need help for this question?

[lethal steeple]

Im still trying to factor it

sorry i mean cancel out terms

[crude hearth]

no worries
Do you recognize the following equality?
a^2 - b^2 = (a-b)(a+b)

[lethal steeple]

yes

[crude hearth]

so, try factoring out the 2 from the denominator

you'll then get x^2 - 49

and x^2 - 49 matches the form a^2 - b^2

because x^2 is a squared term
and 49 can be rewritten as the square of 7

[lethal steeple]

i already have xx+7x divided by 2xx-277 i just dont know what to cancel out

sorry one second

x • x + 7 • x : 2 • x • x - 2 • 7 • 7

[crude hearth]

do you mean (x^2 + 7x)/(2x^2 - 2 x 7^2)?

well, you can factor out the x from x^2 + 7x
which will give you x(x + 7)

and you can factor out the 2 from (2x^2 - 2 x 7^2)
which will give you 2(x^2 - 7^2)

[lethal steeple]

sorry could you explain simpler

[crude hearth]

I'll try to.
So, when we say "we factor out x from (x^2 + 7x)", it's equivalent to TAKING the "x" out from the parenthesis, and putting it outside of it. In other words, we divide the expression (in this case, (x^2 + 7x)) by x, and put x to the left side of the newly evaluated expression

so, in other words,
to factor out x from (x^2 + 7x) will make the expression look like this:
(x^2 + 7x) = x(x + 7)

we've basically divided (x^2 + 7x) by x
then multiplied it again by x to keep its value intact.

[lethal steeple]

okay i understand

how do i know where to put parenthesis

[crude hearth]

hmm, usually when you want to express the product between two terms
you separate those two terms by a parenthesis

for example, if you want to multiply x by x + 7
you write x(x+7)

[lethal steeple]

so the numerator is x(x+7)x?

[crude hearth]

oh, the numerator is x(x+7)
you don't need to add a second x after the (x+7) term
because when you factored out (x^2 + 7x), all you did was to divide that expression by x one time
So, to "counteract" this division, you only need to multiply (x+7) by x one time

[lethal steeple]

how did i divide it by x one time im not following

[crude hearth]

well, to divide x^2 + 7x by x one time
you divide every term of the expression by x once

so, you divide x^2 by x AND you divide 7x by x

which will give you x AND 7
then connect those two terms with the operator that was between them before you divided those 2 terms (which is the addition operator)

[marsh citrus]

@lethal steeple Has your question been resolved?

[lethal steeple]

does this look about right

[crude hearth]

this looks perfect

now, you have to factor the term (x^2 - 49) by using the difference of squares
(where a^2 - b^2 = (a+b)(a-b))

[lethal steeple]

?

[crude hearth]

you unfortunately cannot factor the term x from (x^2 - 49)

but first, you can try to rewrite 49 as 7^2

[crude hearth]

because not all terms in (x^2 - 49) will result in an integer if they get divided by x (for example, 49/x is not an integer)

[crude hearth]

if you rewrite 49 as 7^2, your term will look like this: (x^2 - 7^2)
doesn't (x^2 - 7x^2) look familiar to the form (a^2 - b^2)?

[lethal steeple]

does ^ mean potency

[crude hearth]

no, ^ is equivalent to exponentiation

so, 7^2 = 7 times 7

or 7^2 can be read out loud as (7 to the power of 2)

[marsh citrus]

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[lethal steeple]

.reopen

[marsh citrus]

✅ Original question: #help-33 message

[lethal steeple]

so its 7 times 2

[crude hearth]

no, 7 times 2 is written as 7 x 2
whereas 7 to the power of 2 is equivalent to 7 time 7

[umbral trail]

u have to find the value of x?

[crude hearth]

we just need to simplify the expression

[umbral trail]

oh

@lethal steeple can u send the ques from book?

[lethal steeple]

yes one second

c)

ignore the stuff inside box its an answer sheet

i also dont understand the stuff written there so

[umbral trail]

ok

[wispy jungle]

i think its wrong inside the box

[lethal steeple]

really

[wispy jungle]

excess x outside the difference of squares

[crude hearth]

you're right

[lethal steeple]

my teacher done fucked me up

[umbral trail]

see in the numerator take x as common and in the denominator take 2 as common

[lethal steeple]

u mean make them equal

[languid flume]

Sup

[crude hearth]

no, i think what @umbral trail meant is to factor x and 2 from the numerator and denominator respectively

[lethal steeple]

i dont understand

[umbral trail]

umm now can u understand?

[lethal steeple]

one second ill writer it down

ok i understand but then how do i get to the answer in the answer sheet

or is it wrong

[cunning fiber]

you don't

,w \frac{x^2+7x}{2x^2-98}, x=5

,w \frac{1}{2x-14}, x=5

[elfin berry]

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[cunning fiber]

they just willed an extra factor of x into existence somehow

[marsh citrus]

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

[lethal steeple]

thats weird

thanks for helping me everyone

.solved

[marsh citrus]

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[chrome sand]

hi, i have a question 🙂
we had to count limes and i did it by using l'hospital rule but you can simply use shortened multiplication formulas. did it do it correctly or not?

[main idol]

!show

[marsh citrus]

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

[gloomy merlin]

This is very unusual. Are they asking for the limit of the numerator only?

[chrome sand]

no, the whole thing

[gloomy merlin]

Okay, then show your work like riemann asked

[raven horizon]

so you want to know how one would solve it without using l'hopital?

[chrome sand]

i know how bc it's not hard but i did it by using l'hospital rule and my math teacher said it's wrong

[wispy jungle]

i just did it in my head 😭

[chrome sand]

i wanna know if you can use l'hospital rule anyway

[wispy jungle]

yeah i think you can

[chrome sand]

why?

i have to have arguments to fight for a better grade lol

[wispy jungle]

do double derivatives work in l'hopitals rule?

[lucid bridge]

teacher have the right to prohibit using it

[quick kindle]

yes as long as the successive steps satisfy the hypothesis

[lucid bridge]

though i dont know, are you checking your own equations? perhaps the last three

[raven horizon]

i mean technically you can but some teachers dont allow the use

[chrome sand]

he said we can use l'hospital if it's needed

[quick kindle]

teachers do tend to have a problem with lhopital though and if u havent covered it yet, it is likely prohibited

[chrome sand]

yes, if the first derivative didn't help

[chrome sand]

it was test from l'hospital rule so...

lovely people, can i or can't i use l'hospital rule?

[quick kindle]

you can

[chrome sand]

why?

[quick kindle]

u have a 0/0 case with differentiability of numerator and denominator

[chrome sand]

yes but you can not have it by using shortened multiplication formulas

[quick kindle]

u asked if you can use lhopital rule

u have the ability to in this case

it will lead to the right answer with sound reasoning

[chrome sand]

yeah, i absolutely agree with you so i'm gonna fight for a better grade. thank you so much lovely people ❤️

.close

[marsh citrus]

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[frail zenith]

i've done a sketch of the area but idk if its right i know its meant to be a double integral but im not sure what

[leaden monolith]

Why have you written the line sqrt(x) on the y axis

What does that even mean

[frail zenith]

yeah the way my course has us sketch the area is confusing i think its just meant to be the limits of the integral for dy

[leaden monolith]

This isn’t right

You need to graph y = sqrt(x)

And figure out which side is the < condition

[frail zenith]

something like this then?

[leaden monolith]

Yes

Exactly like that

That’s perfect

Now there’s an easy way and a hard way to do this area

Can you think of them

[frail zenith]

well it being a double integral is whats throwing me off cause ive only just started them so still trying to wrap my head around it

i need to find where they meet im guessing

[leaden monolith]

Why would it be a double integral

[leaden monolith]

That’s the hard way

[frail zenith]

its what were studying at the moment it wouldnt be just a single integral

[leaden monolith]

Oh I’m wrong you have to find where they intersect either way

This is the easy way

Otherwise you need to do this

You need to separate them because the integrands are different functions

[frail zenith]

ok i think ive got it thank you

.close

[marsh citrus]

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[zinc moat]

YO

[zinc moat]

I NEED HELP

<@&286206848099549185>

[wary kite]

what’s confusing

[zinc moat]

i dont know how im supposed to combine the functions

sorry im kina slow

[wary kite]

what would the zeros of f * g be?

[zinc moat]

(-3,0), (1,0), (-1,0)

?

[wary kite]

yes

so plot those first

now what’s the y intercept of f * g

[zinc moat]

the y intercepts are (0,-3) and (0,1)

[wary kite]

i said of f* g

those are for f and g yes

but i want it for the product

[zinc moat]

ohh

so only for f?

[wary kite]

no

[zinc moat]

how do i get it for the procut

[wary kite]

for the product f(x)g(x) = (x + 3)(x + 1)(x - 1)

find f(0)g(0)

call this new function h(x) = f(x)g(x)

we want the y intercept of h

we just found before that h has zeros at -3, 1, and -1

[wary kite]

i.e. h(0)

[marsh citrus]

@zinc moat Has your question been resolved?

[marsh citrus]

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[azure pasture]

Find the curve such that the initial ordinate of any tangent is less than the abscissa of the point of tangency by 2 units

[marsh citrus]

@azure pasture Has your question been resolved?

[marsh citrus]

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[gritty lintel]

How to prove the SAS theorem for the congruence of triangles?

[stoic saddle]

do you have a written reference for a definition of congruence

gonna have to have that if you want a proof

[gritty lintel]

Two triangles are congruent if they are identical, that is, one can superpose the first triangle over the other exactly.

[stoic saddle]

i think you put an extra "n" in the word "identical". but also what is "superimpose" supposed to mean

that really doesn't sound like a workable definition to me, and i don't want to force you to produce the entire fucking list of axioms

[gritty lintel]

I meant "superpose".

[stoic saddle]

changing the spelling does not answer my question

what, in geometric-axiom terms, is a "super(im)position"?

[marsh citrus]

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[narrow horizon]

Can sin⁶Q + cos⁶Q be written as (sin³Q + cos³Q)³

[stoic saddle]

it can if you're ok with it being wrong

[narrow horizon]

Ohh nvm 😭😭

I needed help with this

[stoic saddle]

,rccw

[stoic saddle]

so you need to prove 2(s^6+c^6)-3(s^4+c^4)+1 = 0

[narrow horizon]

Yesss

[stoic saddle]

you may consider using the sum of cubes identity on the s^6+c^6 bit

so that you can use Pythagoras

[narrow horizon]

Wait so how will we use that sum of cubes identity?

[stoic saddle]

x^3+y^3=(x+y)(x^2-xy+y^2) yeah?

what can you put for x and y so that you cook something with s^6 + c^6?

[narrow horizon]

Waittt (s³)³ + (c³)³??

[wanton slate]

Almost

[elfin berry]

Ishmam

[narrow horizon]

OHHH

(S²)³ + (C²)³ maybe??

[wanton slate]

Yeah exactly

Cuz then it will nicely simplify

[narrow horizon]

Ohhh thanks so muchh

[stoic saddle]

lowercase but yes

[narrow horizon]

Ohhh got it

Thankss

[wanton slate]

I guess you can do the rest on your own

[narrow horizon]

Yuppp

How can I disband this??

[wanton slate]

!done

[marsh citrus]

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

[narrow horizon]

.close

[marsh citrus]

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[marsh citrus]

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[polar maple]

as a complex number

[wanton slate]

you want to simplify this?

[polar maple]

yes

[wanton slate]

so look

[polar maple]

I know it means 2^7 x i^21

and i^2=-1

[wanton slate]

okay so good progress

2^7 is 128 right

[polar maple]

yup

[wanton slate]

we just need the value of i^21

[polar maple]

and I think i^21 should be equal to -1

[wanton slate]

okay let's see

[polar maple]

because it repeats every power of 4

becoming positive

[wanton slate]

you know i^4 = 1

[elfin berry]

Ishmam

[polar maple]

i^20 should be equal to 1

[wanton slate]

yeah

so we are left with only?

[polar maple]

i^21 should be equal to -1

[wanton slate]

you can see here

we are left with i only

[polar maple]

oh -i?

or just i

[wanton slate]

no only i

[wanton slate]

look here

[elfin berry]

Ishmam

[wanton slate]

so what do we get as the final answer?

[polar maple]

128i

[wanton slate]

yeah

exactly

anything else you need help about?

[polar maple]

I get confused because i
is both i and sqrt-1

[elfin berry]

Ishmam

[polar maple]

okay thank you

.close

[marsh citrus]

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[still temple]

I'm having some issues trying to know how to find out answers like these?

[still temple]

Never been too good with this stuff

[marsh citrus]

@still temple Has your question been resolved?

[chrome jacinth]

Hi! Before you answer this question mathematically, can you explain what is happening visually?

As in, describe what the lines look like. (In particular, where do the lines start and where do the lines end?)

I.e how would you describe this?

[still temple]

Uhh one of them start at 4 and end at the end of the line to the left

and same for 5 and end of the line to right

[chrome jacinth]

Good! The line in yellow starts at 4, and ends to the very left ... in saying that, is there an end on the left side?

[still temple]

Well it stops at the arrow but idk if that really ends or not

[chrome jacinth]

That is, does the line stop somewhere to the left, or does it keep going forever?

[chrome jacinth]

What do you think the arrow means?

If you had to guess :)

[still temple]

Infinity?

Prolly not

(my self confidence is horrible lol)

[chrome jacinth]

No no you're right! The arrow means it goes infinitely in that direction.

So,

If the arrow is pointing left, where does the line in yellow stop?

(And remember, the arrow means it is going on infinitely!)

[still temple]

It just doesn't stop if it's going to infinity

[chrome jacinth]

Good.

Because it is pointing left, which direction is it going towards? Negative or positive?

[still temple]

Negative?

[chrome jacinth]

Good.

That's exactly it :)

So

The line in yellow starts at 4, and goes to -infinity.

How would you write this as an inequality?

[still temple]

That is something I do struggle with!

[chrome jacinth]

Hmm well let's take a step back then

[still temple]

okayy

[chrome jacinth]

Suppose I drew this line segment in red here.

Can you tell me where the line segment starts, and where it ends?

[still temple]

uhhh does it start at -5 and then over to 10 or is it 10 over to -5

[chrome jacinth]

Good! (It doesn't matter but usually it is smallest to highest).

So, it starts at -5 and it ends at 10.

Now,

The way we write inequalities representing these situations is given by:
Starting Value < x < Ending Value

[chrome jacinth]

So, based on this formula here, how would you write the above line as an inequality?

[still temple]

Before I do, how do I know what < to use?

Because I KNOW there's > aswell

[chrome jacinth]

Well,

< means less than, while > means greater than.

So if you have something like:
Starting Value < x < Ending Value
What it means is that the "starting value" will be the smallest number in your line, and the "ending value" will be the biggest number in your line.

[chrome jacinth]

So for example,

If I said

2 < x < 4

That means that x can be any number bigger than 2, and it has to be less than 4.

Because 2 < x means 2 is less than x, and x < 4 means x is less than 4.

[still temple]

Idk why but i thought i would do it like 2 < x > 4 idk why

[chrome jacinth]

Here's the problem though - if you say 2 < x > 4, it means that "x is bigger than 2 and x is bigger than 4".

[chrome jacinth]

So that's a little different than what I'm describing here

Like for example,

[still temple]

what actually is the x again?

[chrome jacinth]

x just means "any random number".

So 2 < x < 3 is asking for all the numbers between 2 and 3.

For example, x = 2.1 or x = 2.004 or x = 2.95.

[still temple]

ohh ok

[chrome jacinth]

So, if I go back to this example,

How would you represent this as an inequality?

[still temple]

Hmmm

[chrome jacinth]

Hint :)

Btw, don't overthink it - just think about what the starting value is and what the ending value is.

[still temple]

-5 < x < 10?

wait no

Did i do that correctly?

[chrome jacinth]

Good! That's exactly it.

https://media.discordapp.net/attachments/907022703230345237/1447879720472543242/2025-12-09_20_16_58-Untitled_Diagram_-_draw.io_-_Brave.png?ex=69393a91&is=6937e911&hm=fdec0d4cb9fc28fc2b95373366d14b30d988a02e5d4cabd7503e7e854d2c7a39&=&format=webp&quality=lossless&width=1278&height=584

Oh one thing I should mention though that is pretty important.

Do you see how there's a red filled in dot on top of each of the numbers?

Do you remember what a filled in dot means?

[still temple]

I do not

[chrome jacinth]

Basically, a filled in dot means that you include that endpoint as part of your "random numbers".

So in particular,

The inequality that we have is:
-5 < x < 10
Because we can also include both x = -5 or x = 10 as part of our 'random numbers', that means we have to add an equal sign.
That is,
-5 <= x <= 10

The symbol <= means "less than or equal to".

So if we have -5 <= x, that just means that -5 is less than or equal to a random number that we choose.

[still temple]

Okay

[chrome jacinth]

For example, we could pick x = -5 or x = 0 or x = 2 ... just any number that is bigger than or equal to -5.

If it has an open dot, we can't put the equal sign.

[still temple]

right okay