help-23

[noble knoll]

.close

[safe radish]

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[vestal seal]

pls can someone

[vestal seal]

help me on this

dont use any maths lingo

im not very good at that

[icy lance]

no lingo of the maths

[vestal seal]

nah like

people were telling me

equations that i havent learnt

[icy lance]

what do you know about transformations

[vestal seal]

u guys smart so

[vestal seal]

uh

idk man

this a reflection

all ik

[icy lance]

have you ever seen stuff like this?

oh thats blurry, one min

[vestal seal]

nope

i dont remember how my teacher taught me

but it wasnt this

can u tell me what it is

been stuck on this all day

[icy lance]

its the one that says reflection over x-axis

[vestal seal]

hm

yh

so

what does f mean

[icy lance]

f(x) is the original function, in your case you can consider it to be you have f(x)=x^2-3

[vestal seal]

ok

so

so i have to negative that

cuz -f(x)

@icy lance

[icy lance]

yeah?

[vestal seal]

bro

whereu go

help me

[icy lance]

i wander around

[plucky elk]

!patience

[safe radish]

Please wait patiently, and do not interrupt other channels with your question. Helpers in this server are volunteers, and the server cannot guarantee that someone will be able to help you. By being impatient or begging, you will only turn potential helpers away.

In the meantime, please make sure your channel contains the original question, clearly describes what you have already tried, and states exactly what you are having trouble with. This increases your chances of getting a good response.

@vestal seal Has your question been resolved?

[safe radish]

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[vestal seal]

hello

[vestal seal]

<@&286206848099549185>

[twilit kayak]

Was there a ping in here

[vestal seal]

yh i accidently pinged mods

[twilit kayak]

In the future, please don't delete pings, even if they're accidental!

[vestal seal]

alr

[twilit kayak]

It just makes it harder for us to track down what happened

[lean otter]

do you have a question

[twilit kayak]

Also don't ping helpers before you even ask your question

[hard crest]

also please don't ping mods - or helpers - without yeah that

[vestal seal]

i did its just up there

[twilit kayak]

Oh you meant to reopen

[vestal seal]

i think

new here

[twilit kayak]

What specifically is your question

[vestal seal]

part b

[twilit kayak]

Have you seen this

[vestal seal]

yh

i wanted him to explain it

i asked him if i was right

he didnt asnwer me

..

[twilit kayak]

This is correct

[vestal seal]

so

-f(x) would be

-x^2+3

?

OMG

YOOOOO ITS RIGHT

thanks g

.close

[safe radish]

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[tiny frost]

Im looking for a hint on how to solve this $$3z-i\overline{z}=7-5i$$

[flat frigate]

Totalani

[tiny frost]

so what I did was this: $$3(a+bi)-i(a-bi)=7-5i$$

[flat frigate]

Totalani

[tiny frost]

And then multiplied them in

$$3a+3bi-ai+bi{^2}=7-5i$$ Dont know how to progress from here

[tame raft]

just group it up

real and imaginary parts

[flat frigate]

Totalani

[tame raft]

(3a - b) + i(3b - a) = 7 - 5i

so 3a - b^2 = 7 and 3b - a = -5

[tiny frost]

wait what?

[tame raft]

so

you can take the real part of the LHS

which is just all the things without an i in

[tiny frost]

right so the 3a

[tame raft]

well

i^2 = -1

[tiny frost]

oh right

yea

[tame raft]

so you should simplify bi^2 to -b

[tiny frost]

3a-b

[tame raft]

yeah

so essentially

if the LHS and the RHS are equal

then the real part of the LHS and the real part of the RHS should be equal

and the imaginary part of the LHS and the imaginary part of the RHS should also be equal

so 3a - b should = 7

[tiny frost]

aha

and 3bi-ai=-5i

[tame raft]

yes

so we can divide through by i in fact

3b - a = -5

[tiny frost]

ok you lost me there

[tame raft]

so

3bi-ai=-5i

there's an i in everything here

so if we divide everything by i, that just makes it simpler, it just turns into

3b - a = - 5

[tiny frost]

yea ok

Ok and how do you progress from there? the two variables confuses me

is it a equation system?

3a-b=7 and 3b-a=-5

think i got it

thanks for the help

.close

[safe radish]

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[umbral solar]

guys anyone know if this is the right way?

was the change an increase of 117.02% from two of these periods?

periods in which rate of change changed, projected for 10 years (exponent)

[safe radish]

@umbral solar Has your question been resolved?

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[short steeple]

Can someone explain how he made the scale in radians (non calculator please)

[plucky elk]

180 degrees = pi radians

[short steeple]

What

Yes I know rhat

But jow did he make rhe scape

Scale

To determine the major points

[safe radish]

@short steeple Has your question been resolved?

[obsidian rock]

hi

[plucky elk]

find the major points in degrees, then convert them to radians

[safe radish]

@short steeple Has your question been resolved?

[safe radish]

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[safe radish]

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[zinc zephyr]

Help please!

[safe radish]

Please don't occupy multiple help channels.

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[crisp lodge]

Hello, could someone assist me in finding all of the solutions in intervals of 0* <= 0 < 360*

4 cos^(2)0=1

Could someone explain this and help me with derived trig identities. I'm very lost.

[crisp lodge]

Well, explain derived trig identities, which would help me with this.

[round shore]

this is just algebra

you can assume that cos(theta) is x, and first solve for x

[crisp lodge]

I divided both sides by 4: cos^2(0) = 1/4

cos(0) = +- sqr(1/4)

which got me cos(0) = +- 1/2

so, cos(0) = 1/2 == 0 = 60*

I think that's as far as I have gotten, still don't even know if its correct

[round shore]

that is correct

[severe moon]

can someone help me to do this question pls?

[round shore]

find another channel

anyway

so what other values satisfy cos(theta) = 1/2

[severe moon]

sry

[crisp lodge]

I would only assume 300* because we would do 60* - 360* to find the fourth quardrant

[round shore]

good job

(not done btw)

you have a plus-minus symbol in the equation cos(0) = +- 1/2

[crisp lodge]

yeah, yeah, it's plus and minus. So I would have to solve for the -1/2 of cos(0)

[round shore]

yes

which are?

[crisp lodge]

cos(-1/2) = 120 degrees

and

we would do the same and minus 360* from 120

so 240* would be the other degree

[round shore]

good job

do you think we're done?

[crisp lodge]

don't say there are 8 solutions

[round shore]

there arent lol

we're done

[crisp lodge]

jesus, giving me a heart attack

[round shore]

mb

[crisp lodge]

I mean, there is another question similar to this if you'd like to help me with that one

[round shore]

sure

[crisp lodge]

It's this one. Give me a moment and I will show you what I was thinking

Yeah, actually. I have no clue how to start this since I don't understand the trig derivites well enough

What I would assume is we'd have to get zero alone

so 2x^(2)-x=0

[round shore]

are you sure?

[crisp lodge]

No. I'm not sure.

[round shore]

your method is correc

your equation, not so much

so assuming that sin(theta) = x, we would have (complete the statement)

[crisp lodge]

I think I'm still confused, the only thing I was going to do was rearrange the equation and plug it into a quadratic formula

[round shore]

yes

that is right

but $2x^2 - x = 0$

[flat frigate]

Dork9399

[round shore]

is the incorrect equation

lets compare

$2sin^2(x) - sin(x) = 1$

[flat frigate]

Dork9399

[round shore]

and the above equation

[crisp lodge]

wait, would we have to something with

combining the two sins to get like (sin^2)^2

or something kinda like that

[round shore]

hint: look at the right hand side of both equations

[crisp lodge]

or would we instead be adding sin(x) to the other side

[round shore]

what are some differences

[crisp lodge]

the sin(x) is a negative compared to the positive

[round shore]

wait

[crisp lodge]

is that what you're asking?

[round shore]

i said right hand side

what is the right side in the first equation

[crisp lodge]

the right side is really just the 1. it really is just sin(x)=1?

[round shore]

and in the 2nd equation?

anyway

i feel ive spent too long on making 1 point

in your equation in terms of x, you completely ignore the 1 on the right hand side

[crisp lodge]

I'm still a bit confused. I assumed since 2sin^2(x)-sin(x) woudl just become sin(x)

sorry

[round shore]

wait explain your logic

on how you simplified to sin(x)

[crisp lodge]

you can put it into the quadractic formula.

sin(x) = (-b+-sqr(b^(2)-4ac))/(2a)

where the a = 2, be -1 (sin(x)) and c=-1 where you get from moving the 1 to the left

this gave me basically sin(x)=(1+3)/4 which is just 4/4 or 1

so making sin(x)=1, but there would also have to be a negative value

[round shore]

what happened to the plus-minus sign

[crisp lodge]

so sin(x) would be (1-3)/4= -2/4 = -1/2

[round shore]

yes

[crisp lodge]

where i did the plus in minus by have two solutions

which was those two

[round shore]

yes

[crisp lodge]

sorry, have I been doing this wrong

[round shore]

now solving for theta?

[crisp lodge]

kinda just do the same thing as before. sin(x)=1 so sin^(-1)(1)=90*

would we also minus that by 360

so it would give me 270*

[round shore]

wait

for cos, that will work

but for sin, it will negate the value

for example

sin(90) = 1, but sin(360-90) = sin(270) = -1, not 1

[crisp lodge]

and we'd want it to be the same value as sin(x) which is one

alright, so that would only give use the value of 90*

[round shore]

ys

[crisp lodge]

as for the other one would it be -30*

and 210 which is both equal to -.5

[round shore]

in the question it asks for an answer in the range 0 to 360

-30 isnt in that range

but other than that

good job

[crisp lodge]

so we wouldn't have -30 in the question, instead only have 90, 270, and 210

since they are the only ones that are within the range. I getcha

[round shore]

wait

what is -30?

[crisp lodge]

-30 degrees. You solve it from sin^(-1)(-1/2)

[round shore]

yes i know that

but when you conv to a positive angle

what does it become

hint: it isnt 270

[crisp lodge]

I thought it was 210*, it could also be 330*

since -30+ 360 = 330 which makes sense. I just remember hearing 210 for an equation like this

[round shore]

yes

[crisp lodge]

actually, it is 330. It would still give us -.5

[round shore]

210 and 330 work

270 does not

so your solutions are 90, 210,and 330

[crisp lodge]

I see, I will attempt another question and see if I get it right. If I have anymore questions could I come back to you

unless you just wanna wait here I guess

don't know if you're busy

[round shore]

i might stay

idrk

[crisp lodge]

I gotcha, I just get confused when you end up getting questions like sin(x) cos(x) = sin(x)

but I'm going to attempt it first

Here's what I'm thinking. Since both sides have sin(x) in common we can divide by sin(x)

[round shore]

do you want my feedback

[crisp lodge]

So, (sin(x)*cos(x))/sin(x)=sin(x)/sin(x)

sure

never hurts

[round shore]

can you divide by 0

[crisp lodge]

you cannot since it is undefined

but its not specifically defined

I mean it could be zero

[round shore]

and what happens when it is 0?

[crisp lodge]

oh wait, if it's zero that would make the equation one

[round shore]

so theres one solution

[crisp lodge]

so cos(0)=1

[round shore]

and sin(0) = 0

[crisp lodge]

correct; however, is it just assumed to be zero if the equation is not specifically defined

I'm just confused where the 0 really came from

[round shore]

what happens when we take sin to be 0?

[crisp lodge]

if we take sin to be zero, it will always be zero.

So the the one solution would just be 0 degrees

[round shore]

yes

so when you see a common factor on both sides

first see what happens when its 0

and then divide

[crisp lodge]

oh. I typically just assume it's one and call it a day. Also, supposedly these actually two solutions

0 and 180

do you know how they got the 180?

[round shore]

what are the solutions to sin(x) = 0

[crisp lodge]

sin(x) would be 0, 180, 360, 540

and so one

on*

right

[round shore]

yes

so there you have the solutions

anything else?

[crisp lodge]

thats all, I'm going to attempt it again and see if I get it. I appreciate the help!

[round shore]

do you want me to stay?

[crisp lodge]

I mean if you're not busy. I need to solve 3 questions till I'm able to finally able to take the review which gives me feedback on how to do the questions

[round shore]

ight ill stay

[crisp lodge]

don't understand why the lecture notes cover different questions and such. So it's been pretty painful

I appreciate it!

Here's the one I'm working on if you'd like to see what I'm thinking

Alright, since we know 3tan^(2)(0)=1 ... we can divide both sides by three to get rid of the 3. So, tan^(2)(0) = 1/3

get rid of the sqr by sqr'ing both sindes so

tan(0) = sqr(1/3)
tan(0) = -sqr(1/3)

so we would get the answer tan(0) = 30* and -30* but we cannot have -30*

so we can do -30+360 which is 330*

[round shore]

are there other solutions to tan(0) = sqrt (1/3)?

[crisp lodge]

I would assume.

We could do 30 + 180, to get the opposite side of 30 degrees on the circle which would be 210

is that correct?

[round shore]

yes

so you have four sol

[crisp lodge]

so 30*, 150*, 210* and 330*?

we wouldn't use the -30 since it's not within the range

would that correct?

[round shore]

yes

[crisp lodge]

just double checking! Here's the next one.

alright, similarly I think we can start this one by simplifying it.

so we'd get 7sin^(2)(0) + 3sin(0)-5= 0 (Won't show the whole process for times sakes)

we can now put the equation into a quadratic formula

wiat

this doesnt seem right

[round shore]

is it 7sin^2(0)

[crisp lodge]

cannot be, let me try again

give me a moment

Actually, would it be 2sin^(2)(0)+sin(0)+1=0?

that looks a bit more correct

[round shore]

you were closer before

[crisp lodge]

Don't know why I'm struggling so much, let me give it another attempt, I'll put what I have so you can actually see

[round shore]

just send me a pic of your work

[crisp lodge]

gotcha

It could actually just be -6sin^2(0)+1 = 3sin(0)+4

I divided it by 3 which is why it is seemingly so small. I'll check again

No, I'm definitly doing it wrong, redoing it would only give me -3-3sin(0)+6sin^2(0), could you show me what I should be doing

[round shore]

srry i gtg

[crisp lodge]

all good man. I appreciate the help you've already provided. Take care!

[round shore]

you too

[safe radish]

@crisp lodge Has your question been resolved?

[crisp lodge]

.close

[safe radish]

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[alpine igloo]

R -> S is a curve

[alpine igloo]

what is Rn -> R

like what are the names for these

N -> S is a series

something like that

[safe radish]

@alpine igloo Has your question been resolved?

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[atomic zodiac]

How do I know whether to use the chain rule or power rule for finding the derivative for this equation?

[split fulcrum]

Ur differentiating?

[atomic zodiac]

yes

[split fulcrum]

Do you see any multiplication of function here

[atomic zodiac]

no

[split fulcrum]

So then you don’t use product

[atomic zodiac]

okay

thx

[split fulcrum]

This will be chain

Because it’s a composite function

[atomic zodiac]

gotchu

thx

.close

[safe radish]

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[manic jackal]

is a continuous function whose domain is a subset of real numbers (my example here is the square root function) continuous at the boundary of its domain? like is the square root continuous when its input is 0?

[safe radish]

@manic jackal Has your question been resolved?

[manic jackal]

eh whatever

.close

[safe radish]

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[lunar rampart]

help

[safe radish]

Please don't occupy multiple help channels.

[lunar rampart]

<@&286206848099549185>

[icy lance]

if youre done with your other channel, go close it

[lunar rampart]

im not, no one answered me

or should i js post multiple problems in one channel

[safe radish]

@lunar rampart Has your question been resolved?

[safe radish]

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[prisma quartz]

Please help, I watched video on how to add the irrationals. Maybe im doing it wrong, or probably i just do not know

[prisma quartz]

-3i -(3 *5i)

so maybe its -3-8i

[safe radish]

@prisma quartz Has your question been resolved?

[prisma quartz]

i may sleep soon

[ebon fiber]

3*5i = 3 *5*i = 15*i

[prisma quartz]

Oh, ok makes sense thank u

.close

[safe radish]

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[devout shale]

I'm working on this question

and just not quite understanding one of the mass points

So just from labelling A with mass 1

We get all of this information about other masses in our triangle (and E should be labeled X in the diagram)

but I'm wondering now how to deduce the mass at N ?

because we don't have information about the ratio of distances between B and N and C and N

Like I want to say it is just 3 but I don't quite understand the logic there.

<@&286206848099549185>

[safe radish]

@devout shale Has your question been resolved?

[devout shale]

@thorn slate ?

[thorn slate]

i'm lost

[thorn slate]

I think I got it?

so as you've noticed, X is given to be the midpoint of B and M

so B=M=2 which impliex X=4

Now we know the ``mass'' of AN, it will be 4

so it must follow that 4 = 1 + m(N), where m(N) denotes the mass at N

now m(N)=3

this can be verified as well by considering BC

[devout shale]

What here

[thorn slate]

m(N) = m(B) + m(C)

[devout shale]

Mass of AN?

[thorn slate]

I mean like

[devout shale]

Why’s it 4?

[thorn slate]

the total

[devout shale]

Why?

[thorn slate]

this statement

[devout shale]

I already know that X=4

It’s labeled as 4 in my diagram

The only point idk how to deduce is N

Because we have masses at A and X but no relation in distance

And masses at B and C but again the same problem

You see what I mean?

[thorn slate]

there is definitely a relation between B and C. X is the midpoint of BC, i.e. BX:XC = 1:1?

which is why the masses are equal

[devout shale]

You’re confusing C and M

X is the midpoint of B and M

Not B and C

[thorn slate]

oh wait yh

:p

[thorn slate]

i guess if we were to use the idea of balancing masses, we would get that 4X = 1A + 3N and 1C + 2B = 3S where X and S are the balancing points of AN and BC resp.

I'm not sure if this implies N=S though

[safe radish]

@devout shale Has your question been resolved?

[devout shale]

@zinc crown

[devout shale]

@hardy lion I realized after our DM that I don’t actually get this part

[hardy lion]

The mass at N is like the shortcut for the mass of B and C

So at the fulcrum, the total mass of thr balance is 3

Its kinda like this

[devout shale]

At X then you mean?

Wait so is X the center of mass of our triangle then

And we’re trying to balance against A?

But A only has mass 1

So why would N have to be 3 to balance against 1

[hardy lion]

Because X is closer to N than A

Also the mass of B is different to A and C

[devout shale]

That’s complicated huh

[safe radish]

@devout shale Has your question been resolved?

[devout shale]

.close

[safe radish]

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[lean otter]

I just can't set this integral up, for question 22

[lean otter]

Tried something but doesn't seem to be the right integral

Can someone please help?

[hollow pagoda]

whats the integral

whats the question?

[lean otter]

This was my first attempt

Question 22(a)

[hollow pagoda]

have you done the integral for y = tanx?

[lean otter]

Sure

Isn't that not the issue here though

The thing is

I don't even need to compute the integral

Just need to know how to set it up

[hollow pagoda]

why u multiply it by 2pi?

[lean otter]

Cylindrical shells method

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

<@&286206848099549185>

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[pulsar grail]

In the drawing the legs of the angle are intersected four lines.Prove that the quadrilateral ABCD is a trapezoid and the quadrilateral PQRS is not.

[mortal sandal]

More info needed

[pulsar grail]

.

[safe radish]

@pulsar grail Has your question been resolved?

[pulsar grail]

Pls help

[safe radish]

@pulsar grail Has your question been resolved?

[safe radish]

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[coarse trail]

how do i get the critical points after getting the derivative?

[coarse trail]

unsure how to get the zeros of -4(sinx)(cosx)

[frigid locust]

-4sinxcosx = -2(2sinxcosx)

[coarse trail]

yes

[frigid locust]

and 2sinxcosx is equal to ...

[coarse trail]

is that a trig identity

[frigid locust]

yeah double angle identity

sin2x = 2sinxcosx

[coarse trail]

oh okay

so -2sin(2x)

[frigid locust]

yeah

[coarse trail]

then what do i do from there

[frigid locust]

-2sin(2x) = 0

[coarse trail]

yeah im confused what to do from there

[frigid locust]

sin2x = 0

2x = sin^-1(0)

2x = 0

x = 0

solution set would be x = 0 or 2pi

whats the difficult thing?

[solemn dock]

Could we say sinx=0 or cosx=0 and find x?

[frigid locust]

let sinx = 0
-4(0)(cosx) = 0
0 = 0

[coarse trail]

im unsure how u went from sin2x=0 to 2x=sin^-(0)

ii may have skipped this lesson

[frigid locust]

take arcsine or sine inverse on both sides to remove sin function

[coarse trail]

does the -2 just go away

[solemn dock]

Yep. It means that x could be equal to 0 or π

[frigid locust]

u can just solve it , why guess

[frigid locust]

-2sin2x = 0
sin2x = 0/-2
sin2x = 0

[coarse trail]

oh okay

okay

so i got x=0

[frigid locust]

and 2pi

[coarse trail]

is 2pi from the unit circle

[frigid locust]

eh kinda , sine is a periodic function

and its value repeats after 2pi

[coarse trail]

so how did u know it was also 2pi

doesnt that mean 4pi is a solution too

[frigid locust]

yeah

[coarse trail]

oh okay

so because 2pi is in the interval we only count that

[frigid locust]

yes

[coarse trail]

so id get the y values or pi/4, 0, and 2pi

lemme do that now

[frigid locust]

u dont get pi/4

[coarse trail]

well

as an endpoint

dont i need to test it

[frigid locust]

no

[coarse trail]

for the original problem

[frigid locust]

cuz u solved for x

[coarse trail]

for max and mins

[frigid locust]

idk , if u want to u can

the answer would be wrong

cuz pi/4 is wrong

[coarse trail]

oh ok

im just gonna test it cuz our prof said to do so

thanks for helping tho

[frigid locust]

np

[safe radish]

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[wooden oyster]

can i just write it up like so and solve reduce it to echelon

<@&286206848099549185>

[safe radish]

@wooden oyster Has your question been resolved?

[wooden oyster]

<@&286206848099549185>

[final halo]

You want to transpose it

Columns should be the coordinates in a basis

If you are planning on row reducing

[wooden oyster]

havnet i done that

?

[final halo]

No you've done rows

[safe radish]

@wooden oyster Has your question been resolved?

[wooden oyster]

so how should i write it up instead

like so ?

[final halo]

Yeah

[safe radish]

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[uncut frigate]

Given $A=\begin{pmatrix}0&-1\-2&-2\end{pmatrix}$, find a matrix $P$ such that $PAP^{-1}=J$ is in canonical form $\begin{pmatrix}\lambda_1&0\0&\lambda_2\end{pmatrix}$, where $\lambda_i$ are the eigenvalues of $A$.

[flat frigate]

jsidind810

[uncut frigate]

Eigenvalues are $-1\pm\sqrt 3$

[flat frigate]

jsidind810

[hearty egret]

what are eigen vectors ?

[uncut frigate]

please leave

[hearty egret]

?

[uncut frigate]

can someone pls help me

pls

[twin jetty]

everg is trying to help you

[uncut frigate]

i dont think he is

he doesnt know what an eigenvector is

[raven heart]

or maybe he asked you what the eigenvectors were in this question

[hearty egret]

yes i know ... how can i know the word "eigenvector" if not?

[uncut frigate]

an evec corresponding to an eigenval $\lambda$ is a vector $v$ such that $Av=\lambda v$

[flat frigate]

jsidind810

[hearty egret]

I wasn t clear, i would say: what are A's eigenvectors ?

[uncut frigate]

for $\lambda_1=-1+\sqrt3$ i got $\begin{pmatrix}s\-(\sqrt3-1)s\end{pmatrix}$

[flat frigate]

jsidind810

[uncut frigate]

for $\lambda_2=-1-\sqrt3$ i got $\begin{pmatrix}t\(1+\sqrt3)t\end{pmatrix}$

[flat frigate]

jsidind810

[hearty egret]

ok if you have the eigenvectors, say $v_1, v_2$, the hard part is done .. now $P^{-1}$ must be $(v_1|v_2)$

[flat frigate]

everg

[uncut frigate]

why P-1 and not P

and what do i choose for s and t

[hearty egret]

because $P^{-1}$ satisfies that thing $P^{-1}e_i=v_i$

[flat frigate]

everg

[uncut frigate]

what do i choose for s and t

[hearty egret]

they are not important

[uncut frigate]

so s=t=1

[hearty egret]

yes therey are good

now $PAP^{-1}e_i=PAv_i=P(\lambda_iv_i)=\lambda_iPv_i=\lambda_ie_i$

[flat frigate]

everg

[hearty egret]

so $PAP^{-1}$ is diagonal with coefficients $\lambda_1, \lambda_2$

[flat frigate]

everg

[uncut frigate]

No its not

@hearty egret

[hearty egret]

i don t know what was your mistake but i get a diagonal matrix

https://matrixcalc.org/it/#{{1,1},{1-3^0.5,1+3^0.5}}^(-1)*{{0,-1},{-2,-2}}*{{1,1},{1-3^0.5,1+3^0.5}}

[hearty egret]

now i have to leave

[uncut frigate]

what the hell

[hearty egret]

this is the site that i used for matrix computations

[safe radish]

@uncut frigate Has your question been resolved?

[safe radish]

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[light rapids]

Find the number of round-trip commuter rail tickets sold.

1. Thirty times as many round-trip tickets as 12-ride tickets were sold.

2. The total number of tickets sold represented 1440 rides

I tried doing x + 6x = 1440 but it didn't work

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@light rapids Has your question been resolved?

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[acoustic goblet]

Hey guys, i need to proof that every infinite subset of N is equipotent to N itself. Can one look at my proofs and tell me if i got any logicial flaws in it?

[acoustic goblet]

Let X be a subset of N which is infinite. So:

X={x1, x2, x3, ....}, xi in N

Lets create a mapping N -> X
f: N->X
n |--> x_n

this is a mapping bc. all n E N are mapped onto their representatives x_n, which exists because X is infinite (there is no largest i for x_i)

its injective because we can first sort X so that x_(i-1)<x_\i<x_(i+1) bc of the well-ordering principle
and then let a,b be in N, with a!=b --> f(a)=x_a != f(b)=x_b bc. if a!=b -> a>b or a<b -> x_a < x_b or x_b < x_a

its surjective bc. for every x_i we find an n E N for which f(n)=x_i -> i=n

[steep lily]

you're assuming you can write X = {x1, x2, ...}

I don't know what level of formality is expected, so this might pass scrutiny, but to me that's too big of a flaw, unless something has been introduced in your course as a fact that allows you to do it

[acoustic goblet]

wdym i mean if i define x_1:=the first element, x_2:= the second element etc.?

[steep lily]

ok cool so you're using well ordering to define those

I would explicitly write that

[acoustic goblet]

Ahhhh now i get what you mean 😄

Yeah, sure thats good to add

rest is fine?

.close

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[safe radish]

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[thin helm]

i have to prove that some point in the neighbourhood of (0,0) is discontinous

[thin helm]

the original function is $x^2sin(\frac 1 y)$ when $y \neq 0$ and $0$ when $y = 0$

[flat frigate]

omgatriple

[thin helm]

so I assumed point Q = (x', y') where y' =/= 0 is in the neighbourhood

Then, you know point Q' = (x', 0) is also in the neighbourhood

Now, can I show that $$\lim_{(x', y') \to (x', 0)} f(x', y') = \lim_{(x', y') \to (x', 0)} x'^2 \sin (\frac 1 y') = \lim_{y' \to 0} x'^2 \sin (\frac 1 y')$$ and show that doesn't exist and that's enough to prove that point Q' is discontinous?

[flat frigate]

omgatriple

[placid oak]

that should work

discontinuity along a line implies discontinuity

and along that line you basically get a scaled version of the topologist's sine curve, which is discontinuous, so that looks good

[thin helm]

it seems weird to me because i've always used $\lim_{(x,y) \to (a,b)}$ never $\lim_{(a,b) \to (c,d)}$

[flat frigate]

omgatriple

[thin helm]

like it feels wrong to start from non general values

[placid oak]

for showing continuity, sure

but discontinuity you're looking for a counterexample

so that's fine

[thin helm]

ok just to clarify then

it's ok because i am showing a very specific counter example

one which basically happens to be showing that x' doesn't change and y' heads to 0

[placid oak]

yep

[thin helm]

so i'm approaching it along the x axis

[placid oak]

well, parallel to the x axis

[thin helm]

and along that curve, the limit DNE

yes parallel

because x' is something

[placid oak]

but yes, this works because you can pick specific values for a counterexample

if you were trying to show linear continuity, it'd have to hold for all lines

or for total continuity, for all paths

etc.

[thin helm]

and thats when i'd have to have general starting points

[placid oak]

yeah

[thin helm]

instead of a specific opoint

ok great thanks

[placid oak]

and not just starting points, but general paths to the limit point as well (for general continuity)

[thin helm]

yup, that's when squeeze theorem or epsilon delta comes out

thanks a lot, was stuck on that one

.close

[safe radish]

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[barren brook]

is anyone able to help me with this?

[barren brook]

I know about imaginary numbers and how they have conjugates, but not sure how to write it in a+bi

[neat kiln]

Multiply them and see what happens

[barren brook]

so just that?

[neat kiln]

Yeah and i^2 is?

[barren brook]

1?

-1*

[neat kiln]

Yeah

[barren brook]

thank you

.close

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[sonic bluff]

plz help

[neat kiln]

!status

[safe radish]

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

[sonic bluff]

2

[neat kiln]

Show your work

[sonic bluff]

Idk if it’s this way or do I have to set it to another variable to represent it

like U = x/x-8

[delicate sphinx]

That first step is wrong because $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$

[flat frigate]

CaptainNova22

[sonic bluff]

oh ok

[delicate sphinx]

I displayed the wrong thing

I meant $(a+b)^2 \neq a^2 + b^2$

[flat frigate]

CaptainNova22

[delicate sphinx]

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

[flat frigate]

CaptainNova22

[delicate sphinx]

So the left side, you have to FOIL/expand

[sonic bluff]

oh right

how can i foil 3 radical x-8/x

[delicate sphinx]

So you tried to do $\left(\frac{x}{x-8} + 3\sqrt{\frac{x-8}{x}}\right)^2 = 4^2$ then $\left(\frac{x}{x-8}\right)^2 + \left(3\sqrt{\frac{x-8}{x}}\right)^2 = 4^2$

[flat frigate]

CaptainNova22

[delicate sphinx]

But you can't do that because of this

So you need to do $\left(\frac{x}{x-8} + 3\sqrt{\frac{x-8}{x}}\right) \cdot \left(\frac{x}{x-8} + 3\sqrt{\frac{x-8}{x}}\right)= 4^2$ instead

[flat frigate]

CaptainNova22

[sonic bluff]

ohhh

[delicate sphinx]

I forget the square root in the first fraction but you get the idea

[sonic bluff]

yeah]

thanks

thanks alot

.close

[safe radish]

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[junior moss]

Question states:
For what value of m does the equation
(x+1)^3
-------- = m
(x-1)^2

have at least two solutions?

[junior moss]

I'm lost on how to start this

I don't want an answer but I do need help on how to approach this problem

I assume I need to take the derivative of the function

[neat kiln]

Rewrite the equation

[junior moss]

would a photo be better?

[safe radish]

@junior moss Has your question been resolved?

[junior moss]

what do you mean by re-write the equation? like expand it and then cancel?

this question is wonky

every value for y=>13.5 is valid for m if you graph it

[junior moss]

.close

[safe radish]

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[royal ferry]

what have you tried?

[safe radish]

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

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.

[keen canyon]

oh

Im sorry

[delicate sphinx]

Please delete this

[keen canyon]

Im so sorry

Just instant reaction

[nova gulch]

i don’t know how to square it

and then i don’t know what to do after

i don’t know where to start at all tbh

[delicate sphinx]

This is a good start

You need to square both sides

[nova gulch]

but then i get 1= 2x + 5

[delicate sphinx]

Can you show your work?

Because I'm assuming you did $(a + b)^2 = a^2 + b^2$ which is not true

[flat frigate]

CaptainNova22

[delicate sphinx]

Yeah basically what I said, $(a+b)^2 \neq a^2 + b^2$

[flat frigate]

CaptainNova22

[delicate sphinx]

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

[flat frigate]

CaptainNova22

[delicate sphinx]

So you have to FOIL/Expand

[nova gulch]

what is a and b?

[delicate sphinx]

Place holders to represent values

[nova gulch]

well yes

but

in this example

which numbers are a and b?

[delicate sphinx]

So I suggest you square both sides first instead so subtracting sqrt(x)

[nova gulch]

i still don’t understand

[delicate sphinx]

Which part?

[nova gulch]

how to square them because i have no clue what that means anymore because i was taught that to square a square root you just get the numbers that are inside the sqrt

[delicate sphinx]

That's if it's just sqrt(x) by itself

But you have a plus separately the two terms

It's $\sqrt{x} + 1$

[flat frigate]

CaptainNova22

[nova gulch]

so would the first part be x + 1 when squared?

[delicate sphinx]

Not quite

So you're doing $\left(\sqrt{x} + 1 \right)^2$

[flat frigate]

CaptainNova22

[delicate sphinx]

And $\left(\sqrt{x} + 1 \right)^2 = \left(\sqrt{x} + 1 \right) \cdot \left(\sqrt{x} + 1 \right)$

[flat frigate]

CaptainNova22

[delicate sphinx]

So you need to FOIL/expand

[nova gulch]

like that?

[delicate sphinx]

Yep

[nova gulch]

right?

[delicate sphinx]

Yep

Don't forget the other side too

The $\sqrt{x +5}$

[flat frigate]

CaptainNova22

[nova gulch]

how do i do that?

[delicate sphinx]

I mean you just square it

[nova gulch]

that’s as far as i got

can you show me the steps?

[delicate sphinx]

I'm just saying don't forget about it because you don't have the sqrt(x + 5) on the right side of the equal sign

[delicate sphinx]

This part is the process you were taught, square a square root, you get the numbers under the root

The different between $\sqrt{x + 5}$ and $\sqrt{x} +1$ was in $\sqrt{x} +1$ only the x had the root and the 1 had a plus in between

[flat frigate]

CaptainNova22

[delicate sphinx]

So you need to expand it doing this

[delicate sphinx]

Yes

[nova gulch]

what do i do now?

didn’t mean to send that

my bad😭

[delicate sphinx]

How would you isolate x?

[nova gulch]

ummm

do i divide?

[delicate sphinx]

By?

[nova gulch]

2

[delicate sphinx]

Yes

Now you have?

[nova gulch]

x=4

[delicate sphinx]

I guess you went to the end and solved for x

[nova gulch]

yes

so basically if there’s a + between the sqrt and the equals sign then use FOIL?

[delicate sphinx]

Yes

[nova gulch]

and if it’s all inside the sqrt then it’s just what’s inside?

[delicate sphinx]

Pretty much

[nova gulch]

ok thank you so much

i’m going to close chat now and open it for someone else. thank you

.close

[safe radish]

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[radiant depot]

do i have this written out correctly?

[terse lichen]

yep

[radiant depot]

ok so woul d i just do that Pythagorean theorem

to sovle for X

[solemn vault]

You are asked about the angle not side?

[terse lichen]

i mean the questions asked to label the sides

so i assume all sides need to be labelled

[solemn vault]

I that's so then yes

[radiant depot]

right my bad thanks

how do i solve for angle?

[terse lichen]

trig

[fiery merlin]

Use SOH CAH TOA.

Have you done that one before?

[radiant depot]

would it be TOA

[fiery merlin]

Yes, that's right.

You don't have the Hypotenuse, but you do have the other two.

[radiant depot]

ok so that would be Tan right

[fiery merlin]

Yes, that's right.

[radiant depot]

ok so what would i do from there

[fiery merlin]

OK, so tan(theta) = opposite/adjacent.

What's the length of the opposite side?

[radiant depot]

24

[fiery merlin]

What about the adjacent side?

[radiant depot]

tan^-1 (24/15) ?

[fiery merlin]

Right.

[radiant depot]

awesome

so 57.995?

[fiery merlin]

,calc atan(24/15) * 180/pi

[flat frigate]

Result:

57.994616791917

[fiery merlin]

Looks good.

[radiant depot]

sick so thats the final answer right

[fiery merlin]

Yes, that's right.

[radiant depot]

niceee

keeping this chat open ihave to take notes

[fiery merlin]

OK.

@radiant depot You might also want to find out the hypotenuse if you're right about that being what they mean to label it.

For that, you'd use the Pythagorean theorem.

[radiant depot]

yeah i think i know how to do that

can u check this for me

sry its blurry

thats 11 and 5 for the hypotenuse and opposite

[fiery merlin]

,calc acos(5/11) * 180/pi

[flat frigate]

Result:

62.964308210588

[fiery merlin]

What steps did you take?

[radiant depot]

may have used SOH CAH TOA wrong

i also may have done 11/5

instead of 5/11

[fiery merlin]

Oh, OK.

[radiant depot]

wb this one

[fiery merlin]

OK, so you have opposite and hypotenuse.

So, which function is that?

[radiant depot]

SOH

[fiery merlin]

,calc asin(6/14) * 180/pi

[flat frigate]

Result:

25.376933525152

[fiery merlin]

Looks good.

[radiant depot]

oh ok so i got tha tone

and then these last 2

i wasnt too srue on these

i had to kinda guess

[fiery merlin]

OK, so let's look at the first one.

Which two sides do we have out of A, O, and H?

[radiant depot]

just O right?

[fiery merlin]

Well, we have the number for just the adjacent (next to) side.

But we're dealing with three things.

We got the angle and two sides we're dealing with.

Which two sides?

[radiant depot]

ehh

angle B and opposite and adjacent?

[fiery merlin]

Right, so we use TOA.

[radiant depot]

Ohh

[fiery merlin]

Tangent equals Opposite over Adjacent.

tan(74 degrees) = x/3.

[radiant depot]

So if there is a variable still use soa cah toa. got it

[fiery merlin]

Yes, that's right.

Those are about what you're dealing with altogether, not what you know the numbers for.

You're dealing with the numbers and the variable.

[radiant depot]

so when i put it into desmos shouldnt i move the 3? because its x/3

[fiery merlin]

Well, you're solving for x, so you use a bit of algebra to isolate it on one side.

[radiant depot]

Hmm... so what should i do from here? I know its Tan and i have x/3.

[fiery merlin]

OK, so pretend tan(74 degrees) is just another variable, t.

So, you have t = x/3.

How would you use algebra to solve for x?

[radiant depot]

Ok so just 3t=x

[fiery merlin]

Right, so x = 3 tan(74 degrees).

Does it make sense how I did that?

[radiant depot]

Yes i think so

[fiery merlin]

tan(74 degrees) is just a number and a variable like t can be used for a number.

[radiant depot]

would it be tan^-1 or just tan

[fiery merlin]

Oh, just tan.

You're not trying to get the angle.

[radiant depot]

oh ok

[fiery merlin]

The way getting the angle works is you have this:

[radiant depot]

10.462

[fiery merlin]

tan(angle) = o/a

Then you arctan both sides to get angle = arctan(o/a).

[radiant depot]

yeah i get it i think

[fiery merlin]

But you have x = 3 tan(74 degrees). No need to arctan to get x by itself.

[radiant depot]

wait so idid iget the right answer

[fiery merlin]

,calc 3 * tan(74 * pi/180)

[flat frigate]

Result:

10.462243331523

[radiant depot]

nicee

[fiery merlin]

Yes.

[radiant depot]

ok so the last one

soh cah toa <= ignore that

for me lol

[fiery merlin]

What's the number at the bottom? 15 or 115 or 75?

[radiant depot]

so cah

its a 16 sorry

[fiery merlin]

Is the angle 24 degrees?

[radiant depot]

adjecent is 16, the angle is 24 degrees and hypotineus is X

[fiery merlin]

Oh, OK. Yes, you're right about CAH.

[radiant depot]

ok so i would do

t= x/16?

[fiery merlin]

No, first write it out using what CAH stands for.

[radiant depot]

shoot

is it csc

i frogot lol

[fiery merlin]

Cosine equals Adjacent over Hypotenuse.

[radiant depot]

oh yes

[fiery merlin]

So, cos(24 degrees) = 16/x

Now, we can do c = 16/x to make it easier.

How would you solve for x?

[radiant depot]

divide?

[fiery merlin]

OK, so what would you divide both sides by?

[radiant depot]

16

c/16=x

[fiery merlin]

OK, c/16 = 1/x.

No, the x has to stay in the denominator.

[radiant depot]

oh ok

c/16 = 1/x

[fiery merlin]

You can flip both sides, though.

[radiant depot]

now what

[fiery merlin]

OK, so take the reciprocal of both sides.

[radiant depot]

uhhh what that mean again

[fiery merlin]

It means change the fractions so that the bottom and top change places.

Like 2/3 -> 3/2.

[radiant depot]

ok so 16/c = x/1

[fiery merlin]

Right.

So, x = 16/c.