help-10

[unreal musk]

Do watch something on how to solve simultaneous equations

[astral ivy]

ok!

When you couldn't find any videos:

[unreal musk]

you couldn't?

Did you try e.g. this one? https://youtu.be/clY_pbtBneA?feature=shared

[astral ivy]

Nope

[unreal musk]

As in you didn't try it?

[astral ivy]

Yeah because I did not see it lol.

xd

When I searched up Simultaneous Equations!!

I will do it tomorrow because I need to sleep soon. c>

[unreal musk]

Aww, that's all fair enough hope you rest well!

[obtuse pebble]

@astral ivy Has your question been resolved?

[obtuse pebble]

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[stoic merlin]

Not sure if i simplified correctly

Can someone pls just verify

[twilit thicket]

you can't do that

$\frac{x-3}{3}=\frac{x}{3}-1$

[warm shale]

CH3

[twilit thicket]

just do this $\frac{1}{3}\int_{ }^{ }\frac{x-3}{u^{2}}du$

[warm shale]

CH3

[stoic merlin]

I see thank you

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[dense schooner]

Do the steps im taking to solve this question make sense/are allowed

[unreal musk]

One comment-

[dense schooner]

omg hi chartbit

[unreal musk]

Should be (i/n)^2

But otherwise, looks all good to me

[dense schooner]

why is that

is it because i have to divide every term in the equation by n^2

even if it doesnt actually contain n?

ohhhhhh

[unreal musk]

If you divided numerator and denominator you’d have to

[dense schooner]

ok youre right

[unreal musk]

Also remember that xi = i/n so you want that to somehow appear

[dense schooner]

plus i have xi defined as i/n

omg ok

ok i thought i was being slick with my slightly illegal math

unfortunately not 😔

[unreal musk]

you pretty much got the answer anyway in the end when you make two mistakes that cancel each other out

[dense schooner]

hehehe

yes i got the right answer but wanted to clarify if the steps were correct in case a short answer like this comes up on exam

anyways tysm chartbit, goodnight

[unreal musk]

Goodnight to you too

[dense schooner]

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[timid silo]

Real quick question. For the comparison test (if there's a larger function that's convergent, then the function is convergent) does it also work if there's a smaller function that's convergent?

[timid silo]

Or does it have to be larger?

[kind hawk]

has to be larger

[timid silo]

Damn

[kind hawk]

there is always a smaller function which is convergent, the zero function

but if there is a smaller function which is divergent, then your original function is also divergent

[timid silo]

Divergent function can't be larger either?

[kind hawk]

there is always a larger function which is divergent

[timid silo]

Okay ty

[kind hawk]

it doesnt mean anything

[timid silo]

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[wild shoal]

Its a chem question but can anyone please help me with b because I don't really know what to do.

[narrow rain]

First question use this table

[wild shoal]

yeah I was really struggling with part b I don't know if the start of my working is even right

[narrow rain]

Can I ask what level is this first so I know whether I can be of any help because I studied Chemistry in uni but it wasn't very advanced

[wild shoal]

I am in high school

[narrow rain]

Oh ok then I can help

Well... You should find which is Limiting reagent. You find that by comparing solubility in water. NaCl is soluble in water as his Ksp is 37.6 >> to the Ksp of AgCl that was giving in the exercice (1.77x10^-10). So NaCl is the Limiting reagent. From there you get Xmax i.e the Amount of substance of NaCl, calculate the concentrations from that.

I hope I was of any help

[wild shoal]

yeah I havent done this yet lol, I appreciate the help but Ill speak to my teacher tmrw because there are no solutions. thanks again

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[little steeple]

I’ve done all parts but I’m just having an issue with (e)

I’ve shown that b=1 for i greater than or equal to 2

But I’m not sure how to say that the i=1 case will also give b as the identity

This is what i have done so far

This is group theory

[obtuse pebble]

@little steeple Has your question been resolved?

[smoky vigil]

look what happens when you cube the case when i=1 :)

@little steeple

$(a^{2}ba^{-2})^{3}=(b^{3})^{3}$

[warm shale]

Moosey

[little steeple]

Uhh

[smoky vigil]

you can solve for b^3?

[little steeple]

Yes

Wait no

Oh

[smoky vigil]

you see?

[little steeple]

Hmm

[smoky vigil]

:)

but look what we cubed

b^3

[little steeple]

Which gives us b^27

[smoky vigil]

uhhh

[little steeple]

And we can split that

[smoky vigil]

noooo

[little steeple]

Oh

OH

ON THE LEFT

you can get a^2b^3a^-2

[smoky vigil]

mhm

[little steeple]

And the b^27 is just 1

[smoky vigil]

so a^2b^3a^-2=b^9=1

[little steeple]

Ohhhhh okay gotchu gotchu

That makes sense

Thank you!

[smoky vigil]

uhhh

b^27 doesn't show up anywhere

[little steeple]

On the right side you did

(B^3)^3

[smoky vigil]

3*3

not 3^3

[little steeple]

Ah

So it’s just b^9

[smoky vigil]

which is 1

by our definition

[little steeple]

Yeeaaaa

Gotchu

[smoky vigil]

so we can solve like we did in i=2 case to get

[little steeple]

I’m mixing it up with the proof for the previous part

[smoky vigil]

b^3

=1

[little steeple]

Okay nice lemme try writing this down

[smoky vigil]

but also note we had a^2ba^-2=b^3 originally :)

[little steeple]

Yea I’m still in the mindset of the previous proof

[obtuse pebble]

@little steeple Has your question been resolved?

[little steeple]

Thank you for your help @smoky vigil !

[obtuse pebble]

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[fierce vale]

u need to convert each given statement into an equation

then solve the system of eqs

[warm shale]

Adam Chebil

For example, the first statement means : $\\ g(x)=ax^3+bx^2+ax+c$  ;  $(a,b,c)\in \mathbb{R}^3$

[dire lark]

This was what I had been doing

[fierce vale]

yeah so now g(x) becomes:

$g(x)=ax^3+bx^2+ax$

[warm shale]

Adam Chebil

[fierce vale]

since d=0

do u know what a stationary point is ??

[dire lark]

Yes

Where the gradient is zero

[fierce vale]

yeah so u need to calculate the g'(x)

g'(2) = 0 since there's a stationary point at (2,9)

[dire lark]

Would I put (2,9) into the cubic and solve simultaneously?

[fierce vale]

$g'(x)=3ax^2+2bx+a \ g'(2)=0 \ \Longrightarrow 13a+4b=0$

[warm shale]

Adam Chebil

[fierce vale]

yeah g(2) = 9

u'll get a system of 2 equations ( with 2 unknowns)

[dire lark]

👍🏻got it thanks

[fierce vale]

np

[dire lark]

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[fierce vale]

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[vivid fjord]

im looking how to get j1*i0

the professor says its cos (180-45) = cos(45)

but how to look at this picture to figure that out?

[obtuse pebble]

@vivid fjord Has your question been resolved?

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@vivid fjord Has your question been resolved?

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[severe sonnet]

is there a way to find the pattern ?

[obtuse pebble]

@severe sonnet Has your question been resolved?

[cyan sinew]

hmm

[obtuse pebble]

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@cosmic veldt Has your question been resolved?

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@cosmic veldt Has your question been resolved?

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[wide vigil]

i need help on part B

/help

plds help

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[warm trench]

How do i solve these two equation

[pulsar quarry]

taking the lcd and then cross multiplying might help

[warm trench]

So multiply by 12?

[pulsar quarry]

what?

no

[warm trench]

Huh?

Ohhh least common denominator

So multiply by 2?

[pulsar quarry]

no

6

as lcd is 6 not 2

[warm trench]

Huh

[pulsar quarry]

yeaa

lcd is 6

not 2

look again

[warm trench]

Alright .

[pulsar quarry]

yea so you multiply both sides by 6

[warm trench]

Ohh alright

So multipky both sides by 2

Sorry 6

Im saying 2 again🤦‍♀️

Oli wait let me multiply

So noth of the fraction i have to multiply by 6?

After that what do we hv to do

[pulsar quarry]

yea

[pulsar quarry]

what do you have now

[warm trench]

=72

Forgot to write that

Hello?

<@&286206848099549185>

[wide terrace]

is so easy

i am just joking because i'm not very good in math

[warm trench]

Can ya help

[warm trench]

Ohh hehe

<@&286206848099549185>

[pulsar quarry]

uh its wwrong

[warm trench]

So how do i do it

Sp for question 19

3x/2-x/6=12
9x/6-x/6=12
8x/6=12
8x=12×6
8x/8=72/8
X=9

Is that correct?

.close

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[runic bone]

Can someone confirm that the vertical asymptotes are at x+4, x+1

[runic bone]

And the horizontal is at y=0

Just want to make sure

[empty cypress]

,w asymptotes (2x-3)/(x^3-5x+4)

[warm shale]

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Requested by tuzis

[past sand]

"at x+4" doesn't mean anything unless you give x a value

You mean "at x=4"

And yeah you're correct

[empty cypress]

aren't there 3 asymptotes?

[past sand]

Yeah, 2 vertical, 1 horizontal

[empty cypress]

,w x^3-5x+4=0

[warm shale]

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Requested by tuzis

[past sand]

It's x^2 not x^3

[empty cypress]

u are very right

read it as third degree and was pretty sure there were supposed to be 3 vertical

good catch

[obtuse pebble]

@runic bone Has your question been resolved?

[runic bone]

Okay thank you guys

[obtuse pebble]

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[crisp hedge]

no idea how

[fickle oracle]

u know how to derive?

or are you familiar ?

[crisp hedge]

yes ish

u=7x^3+10x^4 v=tan(7x^3+4x^7)

and i derived both of them

but idk where to go next

[obtuse pebble]

@crisp hedge Has your question been resolved?

[unreal musk]

What was the aim of setting u and v like this? Were you planning to use something for that?

[crisp hedge]

I thought this was chain and product rule

[unreal musk]

Yep, fairs how did you think of using them?

[crisp hedge]

I remember my teacher doing something similar

Get U and V

Then U' and V'

[unreal musk]

Yep, so like for product rule, how do you make use of them?

[crisp hedge]

U' x V and V' x U

[unreal musk]

Yep, and then...?

[crisp hedge]

I don't know

[unreal musk]

Well, maybe adding them together might be nice

[crisp hedge]

Oh oops

[unreal musk]

yeah the product rule is (uv)' = u'v + uv' basically

[crisp hedge]

I still don't understand what happens next

[unreal musk]

Well you have everything here, you have u, v, and found u', v' (hopefully correctly!)

Put them in, you have $y = uv$ so $\dv{y}{x} = u'v + v'u$

[warm shale]

@unreal musk

[crisp hedge]

U'= 21x^2+40x^3
V'= sec^2(7x^3+4x^7)?

Not sure about V'

[unreal musk]

u' is fine, v' isn't

You need chain rule for v'

[crisp hedge]

Is the first part of V' right

Just that I'm missing the rest?

[unreal musk]

Yep, you're missing "the rest"

[crisp hedge]

The rest is just 21x^2+40x^3 right?

[crisp hedge]

close

$close

Bro

.close

[obtuse pebble]

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[warm shale]

QuasiStar 超新星

[final thunder]

You have to right multiply B with a matrix with 2 at (1,1) and 1’s along the rest of the diagonal

Column operations are right multiplied and row operations are left multiplied

[obtuse pebble]

@gusty marsh Has your question been resolved?

[final thunder]

BT instead of TB say

[royal mulch]

Do you know about elementary matrices @gusty marsh ?

Yes one row operation away

I think this is a really interesting question I'll probably try to solve it on stream later!

Let's do one of the elementary matrices together

(3) says add row 3 to row 1

Take your 4x4 elementary matrix and do that row operation on it

I want you to see something

I had to write it out though sorry it took me so long

The idea for your question is to find these elementary matrices. If you multiply them on the left of your 4x4 matrix they "act like" a row operation. If you multiply them on the right of your 4x4 matrix they "act like" a column operation.

yeah they are super cool!

see if you can make the other matrices you need for steps (1) to (7) and let me know if you need any help

[royal mulch]

To remove a column, I believe you have to make the column of the elementary matrix zero

then multiply on the right side I think

you'll have to play a bit with that one (that's one of the steps I don't know 100%)

[warm shale]

QuasiStar 超新星

[royal mulch]

I'm about to jump onto stream and will work on it. If you keep the channel open I can double check our work in a few mins

If the channel gets close, feel free to DM me

[royal mulch]

just starting the question now

give me a couple mins

[obtuse pebble]

@gusty marsh Has your question been resolved?

[royal mulch]

OK

so heres what we got

thanks for waiting!

[obtuse pebble]

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[storm kettle]

is this just 0<x<3?

[robust raven]

did you correctly calculate the secod derivative?

[storm kettle]

the seoncd derrivative is linear tho

so i just used the function

[robust raven]

yes

but

[storm kettle]

so there is no concave. upward?

[robust raven]

concavity depends on the secodn derivative

[storm kettle]

if its linear than theres no concavity?

[robust raven]

nwait, second derivative shows >0 or <0

then your second derivative

shows

two intervals

one will show concaivity upward

th eother oen downward

[storm kettle]

-1<x<0 and 0<x<3?

its linear

[robust raven]

y'' = 6x - 4

[storm kettle]

yea

thats linear

[robust raven]

pls now

find a root

of it

draw it on graph

[storm kettle]

oh shet

[robust raven]

x lien only

[storm kettle]

x=-4/6

right

[robust raven]

2/3 yes )

[storm kettle]

so whats the intervals

[robust raven]

you must answer 6x -4 >0

and

6x - 4 < 0

[storm kettle]

so 0<6x-4<0?

that looks wrong

[robust raven]

not togetehr

1. 6x - 4 >0

or

2. 6x - 4 <0

[storm kettle]

wait what

[robust raven]

you wrote together

that wa wrong

[storm kettle]

its 2 intervals?

[robust raven]

let me do it for you, 1 minute plz

$f''\left( x \right)>0\Leftrightarrow 6x - 4>0\Leftrightarrow x\in \left( \frac{2}{3},\infty \right)\\\text{then }f\text{ is strictly concave upward}$

[warm shale]

Joanna Angel

[robust raven]

and

$f''\left( x \right)<0\Leftrightarrow 6x - 4<0\Leftrightarrow x\in \left( -\infty ,\frac{2}{3} \right)\\\text{then }f\text{ is strictly concave downward}$

[warm shale]

Joanna Angel

[storm kettle]

so its -infinity,2/3 and 2/3, infinity?

[robust raven]

but please read my setences

sentences*

[storm kettle]

i dont understand these symbols

[robust raven]

first interval i wrote, gives you concave upward

(2/3, +infinituy) is concave upward

(-infinity, 2/3) is concave downward

[storm kettle]

oh

[robust raven]

is it clear ?

[storm kettle]

so theres only 1 interval?

[robust raven]

yes one interval for concave upward

and one interval for concave downward )

[storm kettle]

do i need to put both

or just upward

[robust raven]

that depends on

what yoru intrcutor wants

your instructor wants concave upward

then you write only interval fo rconcavve upward

(2/3, +infinity))

[storm kettle]

oh alr

[robust raven]

🙂

[storm kettle]

ty

[robust raven]

yw

[storm kettle]

.close

[obtuse pebble]

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[unreal musk]

Might help to factor the numerator and denominator

[obtuse pebble]

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[vague imp]

sos

can someone explain why i subtract (1/4 * 2/5) here?

i understood the adding fraction part

[forest sinew]

its explained in the end there, i think?

do you mean why were subtracting anything at all?

[vague imp]

yeah

[forest sinew]

So I think this is PIE

lemme find a graphic

this one is pretty good

Here's the issue: you count yellow, and you count blue

but green is part of yellow, and its part of blue

so, you counted it twice

[vague imp]

oh gotcha

[forest sinew]

you need to subtract one green to account for that

[vague imp]

thank you so much

[forest sinew]

[vague imp]

😸

[obtuse pebble]

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[hoary flint]

What is the monthly payment rate for a $20,000 loan for 4 years with an annual interest rate of 4.8%?

[obtuse pebble]

@hoary flint Has your question been resolved?

[obtuse pebble]

@hoary flint Has your question been resolved?

[hoary flint]

458.78 is the answer

@hoary flint wow thanks bro

no problem @hoary flint

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[honest flicker]

Doensnt it give me the height

Wtf does the height of the trianglular base even mean

Anwyays I got h≈155.9

[lyric crane]

,calc sqrt(155.9^2+90^2)

[warm shale]

Result:

180.01336061526

[lyric crane]

seems right

[honest flicker]

Wat does that 25 even mean tho

In the word problem

[lyric crane]

doesnt seem relevant to this problem

its the height of the building tho

but you dont need it since its not in the base

[honest flicker]

Wat

If it's the height than isn't the same as the height of the "triangular base"

[sterile vale]

It’s asking you to find the hypotenuse

[lyric crane]

no its not

[honest flicker]

But the hypotenuse is 180

[sterile vale]

Cut the triangle into half, and you can visualize it

[lyric crane]

its asking for the height

h

[sterile vale]

That is the hypotenuse in this situation is it not?

[lyric crane]

no

its a leg

the hypotenuse is 180

[honest flicker]

Hypotenuse is opposite of the right angle

[lyric crane]

correct

[sterile vale]

Damn I just destroyed myself

I’ll take my leave then

[honest flicker]

Ty

.close

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[lament chasm]

very confused rn cuz idk how to write the phase shift

[lament chasm]

i have amp (4.5), vertical translation (13.5) and period (2pi/13.6)

idk how to get the phase shift

ping me if u answer it k bye

[obtuse pebble]

@lament chasm Has your question been resolved?

[cunning burrow]

@lament chasm So we need to write it in the format. To determine the period we need to know k

https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:trig/x2ec2f6f830c9fb89:graphing-sinusoid/v/trig-function-equation?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJBCxIIVXNlckRhdGEiHmthaWRfMTEwMTc1OTQzNjYyNDgyMDk0MTY5Njc5OAwLEghGZWVkYmFjaxiAgNOnzY3uCww&qa_expand_type=answer

The link above discusses curve sketching in detail. If you are still having trouble let me know. Send me ping if you want to verify your answer

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[queen flax]

Basic algebra question here, I want to know how to get from line 1 to line 2

Are both the 1 and the fraction flipped? (reciprocal)

[static marten]

$\frac1{\frac1{1-x}}=\frac{1-x}1$

[warm shale]

chlamydia

[queen flax]

I see, my algebra is pretty weak. Is this flipping 2 fractions in that case?

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

[warm shale]

Jshy

[queen flax]

and that solution being on 1 leaving the 1/1-x

and then flipping that to ge the final answer?

[grand ore]

you have to flip the denominator

[crude comet]

1/(1/(1-x))=1/(1-x)^-1=1*((1-x)^-1)^-1=1 times 1-x = 1-x

[obtuse pebble]

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[grand ore]

yo

[quiet drum]

i thought domain was just the two x intercepts but apparently it’s all the numbers in between is this the same for range also ?

oh hey alpha

[grand ore]

yes

[quiet drum]

okay that’s all i needed to know thanks !

.cloze

.close

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[ember frost]

$x-\sqrt{x^2-1} = 3^{\log_6\frac12}$

[warm shale]

FungusDesu

[ember frost]

i dont know where to start

[oak marlin]

rearrange terms so that you have: x - 3^log6(1/2) = sqrt(x^2 - 1) , and then square both sides. You'll get an x^2 term on both sides of the equation which will cancel, and then it shouldn't be too hard to solve from there.

[ember frost]

so i got $x=\frac{1+\log_6^2{\frac12}}{2 \times 3^\log_6{\frac12}}$

[warm shale]

FungusDesu

so i got $x=\frac{1+\log_6^2{\frac12}}{2 \times 3^\log_6{\frac12}}$
```Compilation error:```! Missing { inserted.
<to be read again> 
                   \tex_let:D 
l.57 ..._6^2{\frac12}}{2 \times 3^\log_6{\frac12}}
                                                  $
A left brace was mandatory here, so I've put one in.
You might want to delete and/or insert some corrections
so that I will find a matching right brace soon.
(If you're confused by all this, try typing `I}' now.)```

[ember frost]

the problem is how do i sinplify this

[timid silo]

what is that random 2 on the log

[ember frost]

log to the power of 2

[grand ore]

huh

oh

log6(1/2)^2

I believe

[timid silo]

i think he means (log6(1/2)^2

i havent check if that is right but if it is i dont think you can simplify it any further

[grand ore]

yeah ^2 of the whole log

lemme try on my side

[grand ore]

wait

what

if you cancel the x^2, there isnt any x's left

[oak marlin]

yes there will be, when you expand (x-3^log6(1/2))^2 you get an x term

[oak marlin]

use the change base formula:

[grand ore]

oh true

lmao

[grand ore]

i forgot all of those properties lol

[obtuse pebble]

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[timid silo]

Uh

[timid silo]

I don’t get C

I got -7/2 for the Gradient at the point (4,-8)

Then I did the negative reciprocal to get the normal

“ Gradient of the normal : “

I need help from there if anyone’s available

[obtuse pebble]

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[echo dawn]

for (a) , i stuck in calculating the determinant of jacobian matrix. What I have calculated is equal to zero which I can't use this theorem.

[echo dawn]

here is the theorem

[oak marlin]

shouldn't 4(x^2)t = 4(1^2)*2 = 8?

Then your determinant would be 8

[echo dawn]

thanks careless

also for part (b) is there any hint for finding ∂s/∂y

I don't know how to find the function s

is this the correct approach?

[obtuse pebble]

@echo dawn Has your question been resolved?

[fathom flicker]

can't read your whole question rn but likely it is multivariable chain rule

[obtuse pebble]

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[cobalt gull]

i can help you

[obtuse pebble]

@torpid agate Has your question been resolved?

[ripe sphinx]

hello

what are u trying to find out

[solemn elk]

Study the variations of f by looking at f'

[ripe sphinx]

I think you are asking about the derivative of the function f(x) = ln(1+e^(2x) - x), which is the rate of change of the function at any point x. To find the derivative, we can use the chain rule and the product rule, which are two common rules for differentiation. The chain rule says that if f(x) = g(h(x)), then f’(x) = g’(h(x)) * h’(x). The product rule says that if f(x) = u(x) * v(x), then f’(x) = u’(x) * v(x) + u(x) * v’(x).

Using these rules, we can find the derivative of f(x) as follows:

f(x) = ln(1+e^(2x) - x) f’(x) = (1 / (1+e^(2x) - x)) * ((1+e^(2x) - x)’ - (x)‘) f’(x) = (1 / (1+e^(2x) - x)) * ((0+2e^(2x) - 1) - 1) f’(x) = (1 / (1+e^(2x) - x)) * (2e^(2x) - 2)

[warm shale]

prog0

prog0

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[keen coral]

The graph G has 12 vertices, and the degrees of its vertices are
8,7,6,6,5,4,4,4,4,3,2,1 .

Prove that G is not bipartite.

[final thunder]

Do you know the handshaking lemma

[undone quartz]

first i'll ask if you understand what a Bipartite graph is

[keen coral]

Yes I am aware of that

@undone quartz @final thunder

[final thunder]

Then just use the lemma…

[keen coral]

Is this right,

To prove that a graph is not bipartite, we need to show that it contains an odd cycle.
Let's examine the degrees of the vertices in the given graph G:
8, 7, 6, 6, 5, 4, 4, 4, 4, 3, 2, 1.
To form a bipartite graph, we need to divide the vertices into two sets, let's call them A and B. The vertices in set A will have edges only to the vertices in set B, and vice versa. In other words, no two vertices in the same set can be connected by an edge.
If G is bipartite, we should be able to divide the vertices into two sets A and B such that there are no edges between the vertices in the same set.
Let's try to do that with the given degrees:
Set A: (8, 6, 4, 4, 4, 2} Set B: (7, 6, 5, 3, 1}
We can see that every vertex in Set A is connected to at least one vertex in Set B, which satisfies the condition for a bipartite graph.
However, if we look closely, we notice that there is an odd cycle in the graph.
The vertices 8, 6, 4, 2 form a cycle of length 4, and the degree of each vertex in this cycle is even. Additionally, vertex 6 is also connected to vertex 7, which creates an odd cycle.
Since we have found an odd cycle in the graph G, we can conclude that G is not bipartite.
Therefore, we have proved that the graph G with 12 vertices and degrees 8, 7, 6, 6, 5, 4, 4, 4, 4, 3, 2, 1 is not bipartite.

[polar fossil]

did u write that yourself?

[keen coral]

Ya practicing for exam tomorrow

[final thunder]

I think this is right as long as you can reference könig’s theorem in your lecture note

Is that even the name idk just the G is bipartite iff it contains no odd cycle

Hm wait how can we infer the existence of odd cycle from the degrees?

[obtuse pebble]

@keen coral Has your question been resolved?

[keen coral]

Never head that 💀😭💀

[obtuse pebble]

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[timid silo]

A satellite dish has a diameter of 100 cm and a depth of 10 cm. Using the formula above, calculate the distance from the bottom of the dish to the focal point where the receiver should be placed.
I just passed the burn point to be (0,1) for The equation for a parabola is given by y = 0,25 ∙ x^2.

[timid silo]

please help me

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[willow crater]

!status

[obtuse pebble]

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

[timid silo]

1. I don't know where to begin.

[willow crater]

The eqn for parabola is given in question?

[timid silo]

yes it is y = 0,25 ∙ x^2.

[willow crater]

Why are u asking me?

[timid silo]

it is y = 0,25 ∙ x^2.

[timid silo]

sorry

[willow crater]

Y=(X^2) 0.25

4y=x^2

[timid silo]

agree

[timid silo]

because 4Y=(X^2) 0.25*4 <=> 4Y=(X^2)

[willow crater]

Yep

[timid silo]

but i am not sure what to do next

[willow crater]

Either the question is wrong or I am wrong

[timid silo]

what do you think

[willow crater]

Since the diameter is 100cm and depth is 10cm , at y=10 the x should be 100/2=50 but it's smth else

X2=4y which means x= 2(√y)=2√10 which is definitely not 50

[timid silo]

the y is 1 for sure

but maybe the x is wrong

[willow crater]

If the parabola eqn is right then the parameters of dish are wrong and if parameters of dish are right then parabola is wrong

Both cases could be possible

[timid silo]

can i do it like this:
4Y=(X^2)
4*1=(X^2)
4=(X^2)
sqr(4)=sqrd(x^2)
x=sqr(4)

[willow crater]

what are u trying to find while doing this

If it's focus then u are wrong

[timid silo]

i am wong

wrong

[willow crater]

The focus of parabola is (0,a) where a is y cordinate of point of intersection of x=2y and parabola

[timid silo]

shoulnt use this formula

[willow crater]

I have not read this formula anywhere so I don't think I am recommended for this

Better ask someone else for help

[timid silo]

who can i ask?

[willow crater]

ping helpers and wait

[timid silo]

<@&286206848099549185>

[willow crater]

What are the x_0 and y_b in this formula

[timid silo]

umm

[willow crater]

Are they the x and y cordinates of focus?

[timid silo]

The reflected ray passes through the two points 𝑃(𝑥0, 𝑎 ∙ 𝑥0
2) and 𝐵(0, 𝑦𝐵) where 𝑦𝐵 is the y-value of the focal point.

[willow crater]

Ok this is out of my league

[timid silo]

.close

[obtuse pebble]

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[robust quail]

how to do this

so far this was my working:
n(n^2+5) by 6

step 1: n = 1, 1(1^2+5) = 6/6 = 1, true

step 2: k(k^2+5) = 6L (l is integer), also k^3+6k = 6L

step 3: n = k+1
(k+1)(k+1^2+5) = 6l
(k+1)((k+1)^2+5) = (k+1)(k^2+2k+6) = k^3+3k^2+8k+6

but i dont how to progress from here

[steel parcel]

try to make it in the form of step 2 so you can replace it with 6L

you know k^3 + 5k = 6L
so 3k^2 + 3k + 6 + 6L
also that 3k^2 = 3(6L - 5k)
18L - 15k + 3k + 6 + 6L

hopefully you can finish it off from there

[robust quail]

yeah, it just becomes 6(3l+2.5k+3k+0.5+l) here and if we let 3l+2.5k+3k+0.5+l be one L, then it should be proved

thanks

[steel parcel]

:3

[robust quail]

.close

[obtuse pebble]

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[timid silo]

I need help with domain and range, i am 15 taking algebra and I just cant grasp it and need a clear understanding of how to do this

[true summit]

what do you understand by domain

[timid silo]

set of values for possible inputs

[true summit]

hm sounds good , can you tell me the cords of the plotted points

[timid silo]

thats where i get lost

[true summit]

coordinates?

[timid silo]

yeah, i just cant seem to understand how to identify them

[true summit]

just see how far the point has travvled from origin in both directions

x and y

for example for the point in second quadrant

for x how much it has travelled from origin

no?

[timid silo]

sorry i was flipping thru my notes, the 2nd quadrant is this guy right?

[true summit]

yep

how much it has travvled for x axis

[timid silo]

i cannot answer that, im not sure

ive never been taught this during my freshmen year - america is weird with math

i cant understand any of this

[true summit]

oh?

np its alright

diff countries have diff education systems , each having their own advantages

anyways

[rapid tusk]

do u know the x and y axis row?

[timid silo]

oui

[rapid tusk]

ok so looking at the x axis what value of x is the point in line with

points are writted as (x,y) so u find the x and y values of the point

we can see that the point is in line with -2 on the x-axis right?

[timid silo]

yes

[rapid tusk]

now how about the y axis?

looking at the vertical y axis what value is the point in line with

[timid silo]

eh, 5?

[rapid tusk]

https://courses.lumenlearning.com/mathforliberalartscorequisite/chapter/identifying-points-on-the-rectangular-coordinate-system/#:~:text=To identify the x-coordinate,order (x%2Cy) . here this might help you

[rapid tusk]

which point are we looking at

[rapid tusk]

lets focus on the one the most left on this picture

it is in line with -2 on the x axist and 3 on the y axis right?

[timid silo]

yes

[rapid tusk]

ok then that point is at (-2, 3)

because points are represented by their x and y values like this, (x,y)

ok now u try the next point to the right of the point we just did

[timid silo]

the point directly on 5?

[rapid tusk]

5 on the x or y axis?

is the point on the y axis?

[timid silo]

y i believe celui-là?

sorry not good english

[rapid tusk]

no problem

u are correct it's y value is 5

what do u think it's x value is

look at the x axis what value is it in line with, the answer is || 0 ||

[timid silo]

0?

[rapid tusk]

so then what is the point

if x = 0 and y = 5

[timid silo]

(0,5)?

[rapid tusk]

yes

[timid silo]

or other way round

[rapid tusk]

correct

[timid silo]

ok

[rapid tusk]

(x,y)

so x first then y

[rapid tusk]

this link has some more practice problems

[rapid tusk]

but lets do 1 more from here

I won't help with this one, what is the next points coordinate?

[timid silo]

(5,1)?

[rapid tusk]

u should get || (1,5) ||

[rapid tusk]

be careful the y value is 5

and the x is 1

so flip them

[timid silo]

(1,5)

[rapid tusk]

correct

do u get how to find the coordinates of a point now?

[timid silo]

oui, merci!

[rapid tusk]

de rien

if u have no more questions u should type .close

[timid silo]

.close

[obtuse pebble]

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[dense tinsel]

I got this completely wrong

How do I solve it? Similar figures type shi

[lone echo]

length on the big triangle are (5+2)/5=7/5=1.4 times those on the small one.
So the area of the big triangle is 1.4^2 times the area of the big one

[dense tinsel]

woah

slow down there partner

can you make it into like a @warm shale diagram thing

and explain why you're doing what

[lone echo]

$\frac{BA}{BP} = \frac{5+2}{5} = k$

[warm shale]

LordFelix

[dense tinsel]

i thought k could be achieved by 2/5

= 0.4

[lone echo]

that would mean a smaller triangle

k=1 means equal size

[dense tinsel]

ooooh

[lone echo]

k<1 means smaller

k>1 means bigger

[dense tinsel]

ok so

we square 1.4?

and multiply 1.96 * 20?

[lone echo]

k is the proportion of lengths

[dense tinsel]

ohhhh

[lone echo]

the proportion of areas would be k squared

[dense tinsel]

and k^2 is the proportion of areas

yeah ok

[lone echo]

and the proportion of volumes would be k^3

[dense tinsel]

what about

4 dimensional objects?

or that's something for another day

whatever

anyway

as I understand it

the answer is 39.2?

[lone echo]

yes

[dense tinsel]

cm^2?

that's

genuinely mind blowing

[lone echo]

you can also do it another way

you got the base and area of the small triangle. So you can get the height

$A = \frac{bh}{2}$

[warm shale]

LordFelix

[dense tinsel]

but it ain't a right triangle

and my teacher says we have to use the similarity factor so this wouldn't help anyway but thanks

[lone echo]

"height" doesnt necessarily mean "side"

[dense tinsel]

so the HEIGHT IS 8

OH MY DAYS

[lone echo]

since A=20cm2, b=5cm, you would have a height h = 8cm

on the small triangle

since they are similar, you also have:

$b_2=kb, h_2=kh$

[warm shale]

LordFelix

[dense tinsel]

PQ || AC, however PQ ≠ AC

[lone echo]

$A_2=\frac{b_2\cdot h_2}{2} = \frac{k\cdot k\cdot b\cdot h}{2} = k^2\cdot A$

[dense tinsel]

nah homie i got 15.68

[warm shale]

LordFelix

[dense tinsel]

wait up tho

OH MY GOD

OH MY GOD

I GOT IT

HOLY SHIT

i am never going to use this system though however because my brain hurt a lot from it

[lone echo]

that's cuz you're not yet used to it

[dense tinsel]

lemme write the first system down though and get this all settled

[dense tinsel]

my teacher also wouldn't accept it on my test corrections

give me a min lil homie

[lone echo]

because they explicitly said "using similarity properties"

there's a difference between being asked the area, and being asked the area following a specific method

[dense tinsel]

so it's 39.6 cm^2

[lone echo]

y

[dense tinsel]

i got another one for you

[dense tinsel]

wdym y its cuz we did 20 * 1.96

next

is hard

[lone echo]

y = yes

[dense tinsel]

oh my bad

anyway

lemme send the next one

give me a sec

it's just this last one I think

so on the left is the edited version

as far as Im concerned

the correct answer is 84.82 cm^3

however my teacher needs to the see the similarity and the k thing

so I need your help with that pratner

[lone echo]

you would start by drawing the similar triangles. The height of the big cone and the radius of the big cone is your big triangle, the ones from the water filled portion are your small triangle

since they are similar triangles, the relation between their sides is constant

[dense tinsel]

alright

but finding the volume of the whole thing would work right

i mean it did for me

[lone echo]

which is what you did in a. Except for you not using units anywhere, which you should.

[dense tinsel]

yeahhh that was a big part of me getting a fucking 77% on this test

anyway continue

[lone echo]

now, your constant would be again the proportion between the small and big triangles. Since you got the heights, you should use those.

$k=\frac{9cm}{3cm}=3$

[warm shale]

LordFelix

[dense tinsel]

yeah ok

k = 3 got it

oh wait

wouldn't it just be 12.6 * 27

since 3^3 = 27

oh wait that's way too big an answer

[lone echo]

gimme a sec

[dense tinsel]

bet

[lone echo]

The amount of water in the cup is not 12.6 cm^3

[dense tinsel]

https://tenor.com/view/woody-woody-toy-story-sora-kh3-kh-gif-27223516

[lone echo]

A cone of diameter 2cm and height 3cm, as it says that is filled with water, has pi cm^3 of volume

[dense tinsel]

wdym

idk

[lone echo]

the teacher used the diameter as radius and rounded to 3 significant figures

[dense tinsel]

bro you're cooking too much

how do i find the entire volume with the similarity 💀

[lone echo]

a cone of radius 2cm and height 3cm would have 12.56637 cm^3 volume

[dense tinsel]

its rounded dude

it's not that deep

[lone echo]

see the bold part

your small cone has a diameter of 2cm, not a radius of 2cm

[dense tinsel]

yeah I know

oh radius

you're saying you found a mistake in the teacher's work?

[lone echo]

so the given volume of water is 4 times as it should be

[lone echo]

not surprised considering your teacher is also not using units properly

[dense tinsel]

bro you're making a michelin star dish

you don't need to cook that much

[lone echo]

i would not accept as valid what they wrote on the left as a correction, since the units are not consistent

[dense tinsel]

just forget the reality and help me with this please

[lone echo]

I ask my 12-year-old students to use units consistently.
I'm not gonna give a pass to a teacher not using them

[dense tinsel]

how old are you?

[lone echo]

31

[dense tinsel]

damn

ncie

but my teacher is using units tho

[lone echo]

not correctly here

[dense tinsel]

you mean the measurements next to the cone itself?

I wrote that

[dense tinsel]

this is my writing, I wrote the volume for a cone thing

[lone echo]

fair enough. Anyways

[dense tinsel]

a 31 year old math teacher using discord is wild

W teacher fr

[lone echo]

assuming that the volume of water given were right (which isnt), you would have the full volume as k^3*V

[dense tinsel]

so it would be 3^3 * 12.6?

[lone echo]

which is 27*12.6cm^3=340.2cm^3

[dense tinsel]

yeahp

[lone echo]

it's a volume. k^3, not k^2

[dense tinsel]

so that's the final volume?

[lone echo]

k^2 is for areas

[dense tinsel]

yeah i know

[dense tinsel]

340.2 is the volume of the entire cone?

[lone echo]

assuming the 12.6ml were right, yes

[dense tinsel]

huh

anyway massive perks to you for finding that mistake

massive respect

dude btw where are you from

[lone echo]

irrelevant

[dense tinsel]

right

ok

im just trying to look for a russian math tutor

[lone echo]

i dont speak russian

[dense tinsel]

oh lmao

what languages do you speak

[lone echo]

also irrelevant

[dense tinsel]

https://tenor.com/view/the-rock-rock-gif-21708339

anyway i thank you tremendously for your help

thanks a lot

I should have just realized this by myself I guess I just needed a push forward

thanks a lot

honestly

.close

[obtuse pebble]

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[obtuse pebble]

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[fluid cargo]

I am not sure what's wrong

nevermind i got it

just entered it wrong

.close

[obtuse pebble]

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[lucid pendant]

i need help with this eqution

equation

<@&286206848099549185>

[obtuse pebble]

@lucid pendant Has your question been resolved?

[obtuse pebble]

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[timid silo]

Can someone see if this is correct?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[edgy matrix]

You cut off the definition of YB at the top, but the working seems right.
Do you have doubts?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[grave crypt]

how do i do 3 e and f

[obtuse pebble]

@grave crypt Has your question been resolved?

[azure anchor]

get them into multiples of special angles. use identities.

135 = 3 * 45 for example

we know 45 is pi/4

3 * pi / 4 = 3pi/4

3pi/4 can also be represented as -5pi/4

then secant is a reciprocal identity of another function

[grave crypt]

where do you get pi /4 from

[azure anchor]

period of cos/sin is 2pi in radians. so 360 = 2pi.

[grave crypt]

we havent even learned radians yet

🙁

[azure anchor]

then you use special triangles.

[grave crypt]

okay how do i use special triagnles in this case

[azure anchor]

secant is the reciprocal identity of what

[grave crypt]

cos

[azure anchor]

alright so we need to find 1/cos(135)

[obtuse pebble]

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[grave crypt]

.reopen

[obtuse pebble]

✅

[grave crypt]

so

-1/sqrt2

because we take 180-135 = 45

?

oh

then do we just flip it to be

-sqrt 2

[azure anchor]

yeah thats correct

[grave crypt]

ahh ok

so what about

csc 270

[azure anchor]

270/45

[grave crypt]

why divide by 45

[azure anchor]

so we can more easily visualize the triangle/unit circle

its 6 * 45, or 3 * 90

so we know we are dealing with a 90 45 45 triangle

[grave crypt]

could you also think of it as 360-270 = 49

90?

[azure anchor]

yes

[grave crypt]

okay so then

1/sin

[azure anchor]

1/sin(-90) or 1/sin(270)

[grave crypt]

so 1 / sqrt 2

[grave crypt]

would it be 1 / srt 2