help-17

[dark kiln]

,calc 1000 (1 - 0.13)(1 - 0.13)(1 - 0.13)(1 - 0.13)(1 - 0.13)

[twin meteor]

Result:

498.4209207

[dark kiln]

the number came from the calculator

i asked it, what number if you multiply by it 5 times becomes multiplying by 1/2

,calc (1/2)^(1/5)

[twin meteor]

Result:

0.87055056329612

[dark kiln]

it's the 5th root of 1/2

[vocal sleet]

@eager elbow Has your question been resolved?

[vocal sleet]

@eager elbow Has your question been resolved?

[eager elbow]

@dark kiln

sorry to ping

24*(o,5) ^ (t/25)

this is correct ?

[dark kiln]

yeah

it's maybe supposed to look different, but it would be equivalent to this

for example 24*2 ^ (−t/25)

[eager elbow]

it supposed to look like that ?

[dark kiln]

no, i don't know

yours looks fine to me, just saying there's many ways to write it

[eager elbow]

i find mine without the 12.94% 0.4998548119289 0.87055056329612 etc

[eager elbow]

i just know that m(25)=12 so i did this

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[humble patrol]

Is 4x + y = 7 as slope intercept form this ?

[formal pond]

Not really

[formal pond]

Y = mx + b is intercept form

[humble patrol]

sorry my photo was taking a long time to load

I mean 4x + 3y = 1

[formal pond]

y= -4x + 7 is slope intercept form

No that is not

Take away the 4x the

Divide by 3

[humble patrol]

sorry I asked the wrong question

[dull bear]

forgot to divide the 1 by 3 as well

[humble patrol]

oh shoot ur right

did I get the -4x/3 part right

[dull bear]

Yea that was fine!

[humble patrol]

oh ok thanks

I thought it was -4x over 3 at first

ok thanks both of you guys for the help I appreciate it 👍👌✅

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[obsidian olive]

i already got my answer and its right

but idk how to get the last part

[dull bear]

Hint: what numbers are you not allowed to divide by?

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[tender glen]

\

[tender glen]

By what is highlighted, are they reffering to surface area?

[visual oracle]

Yes

[tender glen]

thx

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[foggy bolt]

a bit confused on how to get f(x) here

[foggy bolt]

<@&286206848099549185>

[vocal sleet]

@foggy bolt Has your question been resolved?

[foggy bolt]

bro I had xi -_-

thank you

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[light urchin]

how do you identify the volume in this problem? is there a formula

[drifting jackal]

,rotate

[drifting jackal]

Have you tried looking it up?

[light urchin]

Yep I did there were no solutions at all sadly

[drifting jackal]

What do you mean?

I looked up "truncated circular cone" and got a formula

[light urchin]

wait a sec

[jagged cargo]

you already got all the needed parameters

[drifting jackal]

I was trying to get the OP to do it themselves, since this was easily searchable

[jagged cargo]

this is the best i could do, his job now is to do the calculations

[drifting jackal]

I mean I was trying to get the OP to do the research themselves

If you easily found it on google, then the OP can do it too

[jagged cargo]

right, my bad

[light urchin]

I need help with calculating but like is it gonna be 1008π?

Or i might havw calculated it wrong

[drifting jackal]

How did you calculate it?

[light urchin]

1/3 π (3024) = 1008π

[drifting jackal]

And how did you calculate that?

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[plucky epoch]

im not sure if a math server is the correct place but im doing a drawing and i have to convert the real life measurements into a scaled version onto paper but im having issues converting it.

So the scale is 1/4'' = 1'-0''
meaning for every foot in real life it's a quarter inch on paper

So 4 feet would me a full inch (4) quarters

but if i were given 5'-6'' how much would that be on the drawing

[dark kiln]

it's 5.5'

[plucky epoch]

so that would that look like on a ruler

in inches that is

would that mean 5 quater inches and a half

1.375?

[dark kiln]

sure

1 and 3/8

[plucky epoch]

i have no idea how id represent that on a normal ruler

i dont have a scale rule

[dark kiln]

there

there's 16 "degrees" in an inch, i don;t know the name

[cyan shadow]

if they have a name ive never heard it either

[dark kiln]

if 4 feet is 1 inch, then 1 foot is 4 degrees

does the ruler even look like that?

probably yeah

and there's no name for it yeah

no 16

[plucky epoch]

do you know of a calcualtor i could use to make this faster

[cyan shadow]

3 barleycorns lmao

[dark kiln]

uh

change 5 9 and press ctrl-Enter

1 inch 7 unnamed units that i assumed your ruler has

https://tio.run/##y/r/PzU5I1/BRk/BsMhI20DB0EzZyrDIWEFLwdBIWU8BCkwVLP//BwA

added rounding

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[carmine hatch]

I don’t know how to answer without expanding the equation

Finished part a

[pallid zenith]

can you use the chain and product rule?

[carmine hatch]

That’s what I did for part a

[pallid zenith]

youd have to use it here, as well i think

[carmine hatch]

How do I without expanding?

[flat whale]

[stuff]^2

Power rule

[carmine hatch]

Alright thank you I have one more question

For this one I done really understand what it wants me to do

I don’t know what I’m supposed to use y=y(x) for

<@&286206848099549185>

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[woven sundial]

In the opposite figure, a circular disc weighing 4 Newtons resting on a vertical wall and a horizontal ground with the same coefficient of friction was applied to it by a horizontal force tangential to the disc of 15 Newtons until the disc was about to rotate in the direction of the force, so the coefficient of friction = ? (Note that the vertical reaction to the ground is twice the vertical reaction to the wall)

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[vocal sleet]

Please don't occupy multiple help channels.

[worn flare]

That looks like an exam lmao

[coarse zinc]

lol it’s hw

Google form

[worn flare]

Ic

[coarse zinc]

So Ik that 2555 = u1 * 2^(n-1)

But idk what to do next

I’ve been stuck on this for so long

Anyone?

[worn flare]

bro just put it in

[coarse zinc]

Is the answer u1=10

[worn flare]

we might have gotten different answers

[coarse zinc]

If I’m correct n=8

I mean9

[worn flare]

yes

[coarse zinc]

So then 2555 = u1 * 2^8

2555 = u1 * 256

[worn flare]

shouldnt 2555 = U1*[2^9 - 1] ?

[coarse zinc]

Isn’t 2^9-1 = 2^8

[worn flare]

no?

2^9 - 1 = 512 - 1, but 2^8 is 256

[coarse zinc]

Hmm

Oh wait

I see my mistake

It’s (r^n) - (1)

So 512-1

511

2555/511

5

Thanks

[worn flare]

yes

[coarse zinc]

4095 = 3(r^6 - 1) / r-1

4095 = 3r^6 - 3 / r-1

Idk what now

Do I times r-1 by 4095

[dull bear]

Would be a good idea

,calc 4095/3

[twin meteor]

Result:

1365

[dull bear]

Could have also started by dividing by 3 here too

[coarse zinc]

oo

So now I have 4095r - 4092 = 3r^6

Now divide by three

1365r - 1364 = r^6

And then I uhh square root it?

Like the 6 square root

Which gives me r=1

Nice

Wait

No

lol

Hmmm

So -1364 = r^6 - 1365r

[dull bear]

You basically have a polynomial you're trying to find solutions for, may want to move everything to the same side

[coarse zinc]

r^6 - 1365r + 1364 = 0

No idea how to solve this

[dull bear]

Not sure whether you're familiar with the factorisation $x^{n+1} - 1 = (x - 1)(x^n + x^{n-1} + \ldots + x + 1)$ but that possibly might make your life a bit easier

[twin meteor]

@dull bear

[dull bear]

Does look like pain in any case you're probably going to be dealing with a polynomial of degree at least 5

[coarse zinc]

😭

Is there a different way to do this…

We didn’t learn anything like this in class

[dull bear]

There probably is definitely not looking forward to trying to divide that out

I am guessing the aim is that you're to spot something where the solutions are "obvious"

[dull bear]

Making use of this would result in you solving $r^5 + r^4 + r^3 + r^2 + r -1364 = 0$, which I absolutely don't even fancy trying

[twin meteor]

@dull bear

[dull bear]

Guessing that they might want you to use a calculator to find the solutions maybe?

[coarse zinc]

Hmm

Welp

Idk

I guess I’ll just skip it and ask the teacher lol

Thanks for your help

[dull bear]

,w solve 1365(r-1) = (r^6 - 1)

[twin meteor]

Results provided by WolframAlpha

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

Requested by chartbit

[dull bear]

Guessing the r=4 one is supposed to be "obvious"

[coarse zinc]

Huh

[dull bear]

Trying to see what that acutally solves to (r=1 is clearly a solution, the r=4 one is most likely the one you want, the other one doesn't matter too much and looks not to be real)

,calc 4^5 + 4^4 + 4^3 + 4^2 + 4

[twin meteor]

Result:

1364

[dull bear]

Oh, I guess that does work out then

[coarse zinc]

lol

Wow

Hard one

Ty

[dull bear]

definitely worth asking how they expect you to solve that one!

I would cry if they expected you to actually do it "by hand"

[coarse zinc]

💀

Probably not

Yea I’ll ask

lol

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@coarse zinc Has your question been resolved?

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[vocal sleet]

@mellow galleon Has your question been resolved?

[split wind]

Just guessing, is it about eigenvalues and eigenvectors?

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[granite juniper]

how do i simplify or modify nsqrtn
would it be n*n^1/2?

[dull bear]

$n \sqrt{n} = n\cdot n^{\frac12}$ sure, you can keep going

[twin meteor]

@dull bear

[granite juniper]

can i modify the n * n^1/2 more?

[opal gulch]

yeah

[granite juniper]

how:?

[hoary blaze]

think exponent

[granite juniper]

n^1 * n^1/2?

[opal gulch]

base is same so...

powers can

[granite juniper]

are there any others or thats it:?

[opal gulch]

base is same so acc to exponential law you can add the power

[granite juniper]

wait so is it possible to be n^3/2?

[opal gulch]

yep

[granite juniper]

ohh i see thank you!

[opal gulch]

like n*n = n^2

n^1*n^1 = n^(1+1) = n^2

[granite juniper]

.close

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[wild trellis]

İ need help İ have a Q but İ couldnt get it İ dont know how to begin what I need to do here
I understood that I need to find N value

[wild trellis]

@vocal sleet ?

[livid horizon]

Still need help?

[wild trellis]

yeah

[livid horizon]

First try substituting x as -infinity
It is of form infinity/infinity, so we need to simplify
For limits tending to ± infinity, it's generally beneficial if you have the variable in the denominator coz then it would become 0

So try getting x in the denominator in a way that you don't get 0/0 or infinity/infinity again

[wild trellis]

how I can simplify in denominator I have rrot

[livid horizon]

Oh

[wild trellis]

root*

[livid horizon]

By denominator I meant in general

For ex

Let's say you have x/y, you could write it as 1/(y/x) too, over here x is the denominator of another denominator (i didn't know how else to explain this welp)

So you could have denominators within the root As well

Ok well how about you try dividing by x(both numerator and denominator)? You might see my point then

[wild trellis]

I will try and I send you picture how I wrote

[livid horizon]

Sure

[wild trellis]

5 min

so i get that in numeratot i have

1/x - 3

in denominator I have sq root x^2 - 1

like this or in denominator i need sq root x - 1/x

[livid horizon]

This

Denominator is (sqrt (x^2 - 1))/x, if that's what you meant

[wild trellis]

yes

[livid horizon]

Yes

That's right

Now can you think of anything?

[wild trellis]

I have both sides 1/x

i need simplfy ?

[livid horizon]

Wdym

[wild trellis]

like both numerator and denominator

i have 1/x

[livid horizon]

Yeah

Yes we do

Were you thinking about multiplying x or smth?

[wild trellis]

-1/x multiplying 3/ sq root x

like this ?

[livid horizon]

No look
The numerator is done coz we have 1/x in numerator

We need to sort the denominator now

[wild trellis]

`cause in numinator I have 1/x - 3 not 3-1/x did yo get what I say?

hmm

[livid horizon]

Ye i did
But it's alr

[wild trellis]

what about 3 ?

oh

[livid horizon]

Ye leave the numerator as it is

[wild trellis]

so its stay already

[livid horizon]

How can you get a 1/x in the denominator now?

[wild trellis]

i need to change denominator

[livid horizon]

Yes

[wild trellis]

ok

[livid horizon]

the new denominator is sqrt(x^2 - 1)/x

[wild trellis]

yes

I divide another x

so I have sqrt x - 1/x divided by x

[livid horizon]

But that would be within the sqrt

So the overall thing would be
sqrt((x- 1/x)/x)/x

[wild trellis]

yes

[livid horizon]

But not all x terms are in the denominator

Coz it's x- 1/x

[wild trellis]

yes one x is out of aqrt

the only in sqrt its x-1/x

[livid horizon]

Yeah

[wild trellis]

then

[livid horizon]

But if you put x as infinity

It will give another indeterminate value

[wild trellis]

x is not equal to 0

I have number 3 see

[livid horizon]

Where's 3?

[wild trellis]

sqrt((x- 1/x)/x)/x equal to 3

in picture

[livid horizon]

How?

But that's what we've to show

[wild trellis]

.

[livid horizon]

Okok how about this

Can we write x as sqrt(x^2)?

[wild trellis]

yes

we can

[livid horizon]

So the denominator can be written as sqrt(x^2-1)/sqrt(x^2)?

[wild trellis]

what will change then

yes

i got

[livid horizon]

Nice

Well it would be 1/x^2 in the denominator mb, but I think you get the point

[livid horizon]

You can write the entire thing under a single sqrt

[wild trellis]

I dont get this part

sqrt(x^2-1)/sqrt(x^2) in this I got

then what we do

1/x^2 in the denominator mb ???

[livid horizon]

Yes so

[wild trellis]

what is mb ?

[livid horizon]

.

[livid horizon]

My bad

[wild trellis]

oh

[livid horizon]

Yeah

So then you could write it as
sqrt(x^2 + 1)/x^2)

And that would be
sqrt (1+ 1/x^2)

Can you see why?

[wild trellis]

okay okay

you combine into 1 sqrt

[livid horizon]

Yeah

[wild trellis]

then you simplified

[livid horizon]

Yes

[wild trellis]

at the end we get sqrt (1+ 1/x^2)

[livid horizon]

Yes but did you understand how we simplified that part?

[wild trellis]

yes exactly

[livid horizon]

Ok nice

[wild trellis]

sqrt (1+ 1/x^2) but why plus

not minus

[livid horizon]

Oh coz in the q it's x^2 + 1 in denominator

I wrote - by mistake

[livid horizon]

In the original q

[wild trellis]

here we write -

[livid horizon]

Ye that was mistake

[livid horizon]

But anyway
This is the new denominator

[wild trellis]

everythin is plus the

then*

ok

[livid horizon]

For denominator, yes

[wild trellis]

so we have in numinator 1/x - 3

and the new denominator

[livid horizon]

And the numerator was this

Yeaaa

Now put the limits in

[wild trellis]

limit is -infinity?

or what

or I need to put 3

[livid horizon]

Yes

[wild trellis]

how I can do

I only know the numbers for ex I can put any number on limit but what about - infinity

[wild trellis]

?

[livid horizon]

As x tends to - infinity, 1/x tends to 0

[wild trellis]

ok

[livid horizon]

Ye

[wild trellis]

in numinator I have 3

in denominator ?

1 ?

[livid horizon]

Yeah

3/1

[wild trellis]

answer is 3 >

[livid horizon]

Yeah

[wild trellis]

then thats all or what

[livid horizon]

Yeah we proved the limit to be 3

[wild trellis]

so in this Q the answer is 3

[livid horizon]

1-3x/sqrt(x^2 - 1)?

[wild trellis]

but there was N values it was complexity

[livid horizon]

I thought we had to prove that

[wild trellis]

ohhhh

I get it

they give as 3

[livid horizon]

Waut

Yeah

[wild trellis]

we get 3 at the end by proving

but they write e=0.1 and 0.05

what is that?

illustrate definition (0.1) by finding values this one'

thats it or smth need to do

[vocal sleet]

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[vocal sleet]

@young orbit Has your question been resolved?

[vocal sleet]

@young orbit Has your question been resolved?

[lone linden]

.close

[vocal sleet]

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[twin rampart]

Is 880 degrees the same as 90, 160 or 180?

[dry yacht]

How many degrees are there in a circle?

[twin rampart]

360

[dry yacht]

And how many whole circles (360) can you factor out?

[twin rampart]

2

[dry yacht]

So what would you have in rest?

[twin rampart]

180 degrees

[dry yacht]

880-720=180?

[twin rampart]

My bad, 160

[dry yacht]

There's your answer.

[twin rampart]

Ty

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[clever garden]

8x - 5 = 2x + 6

[vocal sleet]

Please don't occupy multiple help channels.

[loud pivot]

Can we get any details?

What is your objective

What have you done so far

Show your work if you've done anything

[wicked sonnet]

Please do not occupy multiple help channels. Keep to one channel per question and close the other one using .close

[loud pivot]

!nosols

[vocal sleet]

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.

[astral drift]

my bad ill delete it

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[finite anvil]

wouldn't the approximation also have I(20,1) in it?

since the linearization formula is f(x,y) = f(a,b) + (x-a)fx(a,b) + (y-b)fy(a,b)

but here they just did (x-a)fx(a,b) + (y-b)fy(a,b) essentially

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<@&286206848099549185>

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[hardy crater]

This is the question, wanted to check if I have left my answer in the correct form since it says exact form.

[hardy crater]

,rotate

[hardy crater]

Ignore the cat

[lucid bane]

,rccw

[hardy crater]

5a is the one

[lucid bane]

yeah, you can just cube both sides to get x :)

[hardy crater]

Alright. Thank you 👍

[lucid bane]

and make sure to plug into the original equation to make sure it satisfies :)

[hardy crater]

For the next one o assumed y=x^2

However one of the roots was negative

And you can’t have a negative root

So I wasn’t too sure on that

[lucid bane]

you can't pull negative out of root, you could technically write that as pm sqrt(5) i and other one is also a plus minus

[hardy crater]

Oh yea, I always forget to include that

Thank you so much 👍👍

[lucid bane]

np :)

[hardy crater]

Have a nice day

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[vast shale]

'

[vast shale]

How do I know which belongs to which?

[cyan shadow]

plot a few points on each, then start regonizing which function forms go to which shapes

to give one of the answers, (A) is a circle, which is just a formula that you know after a while

[vast shale]

can you give me example of you doing this?

[cyan shadow]

sure, lets look at the first graph. It goes through the point (5,2)

Now we can look at the possible equations and see which ones include that point>
x+2y = 4 -> 5 +2(2) = 4 NOT TRUE
...
EQUATION C
2sqrt(x-1)-2y=0
2sqrt(4)-2(2) = 0
2(2) - 2(2) = 0
0 = 0
TRUE

so graph number one is equation C (because (5,2) isnt in any other graph)

if you recognize shapes at all its way easier, i.e. 1 is clearly a sqrt function, 2 is a sideways parabola, 3 is a line, 4 is a correct parabola, 5 is a vertical line, 6 is a circle

if you know the matching equations then you can just do it fast

[vast shale]

tysm

@cyan shadow

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[chilly cave]

is it right, that σ2^-1 is σ5 ?

[mild flower]

doesn't seem like it

[chilly cave]

what would it be then?

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<@&286206848099549185>

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[long gale]

I nee dhelp

[long gale]

I don't get how they intersect

i don't know how to get to the point they intersect at

[pale perch]

they arent parallel

so they must intersect at some point

[long gale]

ik

but idk how to find that point

[pale perch]

set them equal to eachother

x+7=2x-15

since they have the same y coord at their intersection

then solve for the x of said point

[long gale]

would it be (22,29)

??

[pale perch]

seems to be so

[long gale]

ok thank you sir

have a nice day bye

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[mortal geode]

hi this topic is fresh to me so if i don’t understand what you’re trying to say to me it’s because of that but i need help on the problem with the graph bc idk what to do for it lol

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[vast shale]

@vocal sleet

[pale perch]

command recieved

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@median yarrow Has your question been resolved?

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[digital crescent]

how do i solve this simultaneous? im stuck at 3x^2 + 11x + 4

[floral pike]

Substitution

It's just a quadratic

[digital crescent]

i mea i got 3x + 2 = y

i then squared that

[floral pike]

y = 3x - 2

[digital crescent]

nahhhh im acutally clapped, howd i make another sign error with rearranging

tysm, sry 4 wastig ur time lol

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[strong cedar]

I need help integrating this curve

[strong cedar]

My calculator says 3.63897

Which seems more accurate

But i have no clue how it arrived at that number

Yes @vast shale

I realise I could've subtracted the top line from the bottom to get an eq for the area inbetween but my question will stay the same

Or bottom from top sorry

Ok basically

I integrated 2x + 2sin(x/2) to get x^2 - 4cos(x/2). When I enter the boundries, my area comes to -0.36103. My calculator says it should be 3.63897

I have no clue how its 3.63897

I believe it

I just dont know how to get there

I have realised the problem here. I falsely assumed that because its 0 i wont have to do it because it'll amount to 0 anyway

A habit of mine

Yeah it just passed my mind. Thank you anyway

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[glacial leaf]

(25cm- P)^2 = P^2 + 5cm^2
How do I proceed with this?

[heady ibex]

you expand the parenthesis

[glacial leaf]

P = perpendicular

[glacial leaf]

what

[heady ibex]

multiply it out

[glacial leaf]

saying that doesn't help much

[heady ibex]

do you know what to do when you have $(a + b)^2$?

[twin meteor]

Katharine

[glacial leaf]

yep

use algebraic identity

[heady ibex]

you have to get rid of the parenthesis

[glacial leaf]

okay

[glacial leaf]

how do I that's the question

since I saw someone solve this by taking the formula

$(a - b)^2$

[twin meteor]

stuffy

[glacial leaf]

for the first bracket

but not 2nd

[heady ibex]

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

[twin meteor]

Katharine

[heady ibex]

then use the foil

mnemonic

[glacial leaf]

so I was puzzled as to if I should apply the formula in both the sides

[heady ibex]

what do you mean?

[glacial leaf]

well

for the first bracket $(a - b)^2$ should be used

[twin meteor]

stuffy

[glacial leaf]

and it seems for the RHS $(a + b)^2$ could be used

so I'm not sure if I should or not

[twin meteor]

stuffy

[heady ibex]

the right hand side is fine the way it is

there's no brackets there

[vocal sleet]

@glacial leaf Has your question been resolved?

[glacial leaf]

so if there's no brackets i don't need to?

[heady ibex]

no

on the right hand side you have $P^2 + 5 \text{cm}^2$

[twin meteor]

Katharine

[heady ibex]

you don't need to do anything with that

[glacial leaf]

hmm, so even if its written $a^2 + b^2$

[twin meteor]

stuffy

[glacial leaf]

?

[heady ibex]

i'm confused as to what you mean you might have to do

[glacial leaf]

since that can be summed up as $(a + b)^2$

[twin meteor]

stuffy

[heady ibex]

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

[twin meteor]

Katharine

[glacial leaf]

all im confused about is why the algebraic identity applies to once side & not the other, but I may have my answer.

[glacial leaf]

ohhh

why not

[heady ibex]

because you have cross terms

you can write it like this

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

[twin meteor]

Katharine

[glacial leaf]

ok so

no bracket no applying identity?

[heady ibex]

what identity are you talking about?

[glacial leaf]

algebraic ones

that we are discussing here

[heady ibex]

can you write it out

[glacial leaf]

haha

[glacial leaf]

these

[glacial leaf]

.

[heady ibex]

i wouldn't call them identities

they're just multiplying the brackets out

[glacial leaf]

mhmm

so is the answer yes?

[glacial leaf]

.

[heady ibex]

you don't have to multiply out brackets if there are none

🙂

[glacial leaf]

hahaha

[glacial leaf]

thank you for your help 🙏

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[lunar aspen]

can anyone help with this hyperbola?

[faint atlas]

ye

thats smple

@lunar aspen

just find the intersection

divide by 2

and times it by the denominator

as u see the intersection is 1,3

now you can solve it

[lunar aspen]

Not sure if I’m doing it correctly

[faint atlas]

all of that is wrong

[lunar aspen]

yikes

[faint atlas]

a2 + b2 = c"

[lunar aspen]

can you walk me thur it please

[faint atlas]

ok

so

pythagorus therum

what is it

a2 + b2 = c2

right

[lunar aspen]

yes

[faint atlas]

that is why

in life

we need that

to survive

we need that therum

to surivive

to live

[lunar aspen]

following

[faint atlas]

right

now

to the question

it says

the vertices are labeled blue

this is very important

in learning the fundemantles

of surviving in life

and thats how u get the answer

[lunar aspen]

ok...

still a little confused how i would solve it

[faint atlas]

you need to focus on ur academics

this is a year 8 question

should i just give u the answer

[vale nexus]

if you know the standard form equation its a straight forward problem

figure out a, b, h, and k

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@lunar aspen Has your question been resolved?

[lunar aspen]

2

still confused

?

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[autumn orbit]

Hi do you know how to derivate this?

[sweet birch]

very carefully with a bunch of chain rule and product rules

[autumn orbit]

well I tried and I am getting wrong results

[somber badger]

show ur working out

its an ugly derivative

[autumn orbit]

I am getting sin^2(1/x) + x*sin(2x)

[somber badger]

.

[hushed pewter]

How?

[fathom ridge]

That the result

[fathom ridge]

Lol agree

Too much thing in one derivative

Derivative of [ (sin(1/x))² ]= 2(sin(1/x))cos(1/x)(-1/x²)

[hushed pewter]

!nosols

[vocal sleet]

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.

[hushed pewter]

just fyi

[autumn orbit]

that is my work

[fathom ridge]

Now apply the rule (x)'(sin²(1/x))+(x)(sin²(1/x))'
And then replace the derivatives

[fathom ridge]

Btw i don't have helper rule , but alr

[hushed pewter]

The rule applies to anyone who helps.

[fathom ridge]

Ok sir

[hushed pewter]

This is your mistake

Not using chain rule correctly.

[autumn orbit]

I derivate the first and second term whic are both sin(1/x)

[hushed pewter]

That's product rule, not chain rule

[f(g(x))]'=f'(g(x))*g'(x)

[autumn orbit]

ok thanks

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[vast shale]

Hey

[vast shale]

I need help

It is the (x,y,z) three variable linear equation system of equation by elimination method

Idk what to do in this part

@fiery urchin

Nvm

Noone is going to save you but yourself

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[narrow hamlet]

How do we know that ? Is there a way to do this ?

[feral light]

do what

[narrow hamlet]

Assume we do not know the right hand side first

Then how do we find the right hand side

[feral light]

oh

partial fractions?

[narrow hamlet]

But this one is take away

Confused 🤔

[zinc hatch]

if we want to change the right hand side into the left hand side, we want the denominator to eventually look like k(k+1)(k+2)

but, for instance, the 1/k(k+1) term only has denominator k(k+1) : how can we introduce a (k+2) term into the denominator here?

[narrow hamlet]

1/k(k+1) = k+2/k(k+1)(k+2)

[zinc hatch]

yep, and do that for the other term, you should be good

[narrow hamlet]

1/(k+2)=k(k+1)/k(k+1)(k+2)

Do we always use the same trick for all the fractions ?

[zinc hatch]

generally yes, for this type of thing -- we want to make the denominators the same so that you can just add up the things in the numerator

[narrow hamlet]

Okkk thanks

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[tawdry socket]

Solving inverse function. Which is correct?
Or help me prove the mistake

[flat whale]

Where's f(x)

[tawdry socket]

Uhhh

Wait

[tawdry socket]

Using two different method

Alright

Two different methods

Uh wait... I firgot inverse that first

Thank you

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[versed maple]

please help me

i have no idea what those numbers are

somebody please help me

??

[ornate ember]

the opposite? Do you know what they mean by that?

[versed maple]

it's like 15 and -15

[ornate ember]

ok that's what I figured. Opposite is not a good word for that though lol

ok so let's say that we define our two integers a and b

[versed maple]

yes

[ornate ember]

we know their product is some other integer, let's call it p

so we have 1 equation

ab = p

We also know the second integer is increased by 4 and that this multiplication is 16 more than the original product and is negated

so we would have a(b + 4) = -(p + 16)

now we have to equations and we can solve

forgot the 1 lol

[versed maple]

its okay

and then?

[ornate ember]

now we have a system of equations and we can solve for a and b

[versed maple]

i know how the equation works i just don't know what those numbers are i've been trying to find it but i just can't find it

[ornate ember]

wait hold on let me look at this lol

[versed maple]

take your time

[ornate ember]

huh

not as straightforward as I thought lol

ah ok I think I got it now (took a sec)

so we know that we have a product ab.

We know that if we increase b by 4 we get 16 more than the product.
We also know that we get the negation of our original prodcut

so we get two equations out of this

a(b + 4) = ab + 16
a(b + 4) = -ab

Expanding out the first equation, we see the following:
ab + 4a = ab + 16 ---> 4a = 16 --> a = 4

For the second equation, we can plug in our a = 4 we just found:
4(b + 4) = -(4)b ---> 4b + 16 = -4b --> 16 = -8b --> b = -2

So then we get a = 4, b = -2. Let's confirm that this makes sense:
4 * -2 = -8
4 * (-2 + 4) = 4 * 2 = 8
There's a difference of 16 between the two products

so we can confirm it

a = 4
b = -2
🙂

[versed maple]

oh i got it

thank you

you're the best

[ornate ember]

np!

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@versed maple Has your question been resolved?

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[brave ruin]

This is calc 2. So, normally I'm given a value of t to use to get the slope. But in this case I think I'm given just x and y coordinates.

How do we get a t value here to get the slope?

Cause I know the slope for this would be a ratio of two derivatives of the x and y functions

[flat whale]

Set x=-1 and solve for t

Repeat for y=1

[brave ruin]

Now, does that give you two t values ?

[flat whale]

Try it

[vocal sleet]

@brave ruin Has your question been resolved?

[vocal sleet]

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[marble kettle]

2.8.2 only

[hushed pewter]

Is 2.8.2 right?

Is $f^-1(\overline{S})=\overline{f^-1(S)}$?

[twin meteor]

SWR

[hushed pewter]

What kind of spaces are A and B?

@marble kettle, $\overline{S}$ is closure of $S$, right?

[twin meteor]

SWR

[marble kettle]

yes that one!

sorry i didnt see your message

[hushed pewter]

I have a doubt about this exercise

[marble kettle]

why is that?

[hushed pewter]

Let $A=\mathbb{R}\setminus\left{\left(n+\frac{1}{2}\right)\pi :n\in\mathbb{Z}\right}$

[twin meteor]

SWR

[hushed pewter]

And let $B=\mathbb{R}$

[twin meteor]

SWR

[marble kettle]

what is the backslash

[hushed pewter]

We define the function $f :A\to B$ as $f(x)=\tan{x}$

[twin meteor]

SWR

[hushed pewter]

set minus, aka set difference

Basically, A is the domain of the tangent function

and B is the range (all real numbers)

[marble kettle]

when did tan become a thing in all this??

[hushed pewter]

It's an example function to demonstrate why I think this exercise is erroneous

Are you dealing with specific types of functions?

[marble kettle]

i did find an answer online that i believe to be correct but im not 100% sure i understand

https://math.stackexchange.com/questions/3988266/let-f-be-a-function-from-a-to-b-let-s-be-a-subset-of-b-show-that-f−

[hushed pewter]

Oh

[marble kettle]

x exists in inverse compliment S. Based on the definition of compliment there exists a y in compliment S is the same statement as there does not exist a y in compliment s. since f(x) = y then f(x) is not part of s?

[hushed pewter]

$\overline{A}$ is the COMPLEMENT of $A$, not the CLOSURE.

[twin meteor]

SWR

[hushed pewter]

I see

[marble kettle]

i am so lost

i dont even know where to start

[hushed pewter]

Could you share with me your definition of $\overline{A}$

[twin meteor]

SWR

[hushed pewter]

If, say $A={1, 2, 3}$, what is $\overline{A}$?

[twin meteor]

SWR

[marble kettle]

it is basically anything that is not part of A

correct?

[hushed pewter]

Okay yeah

Very unfortunate notiation because that same notation is used to signify closure in analysis and topology courses. So I got quite mixed up there.

Okay, let's start over.

[marble kettle]

Ah I see

[hushed pewter]

What have you determined so far?

[marble kettle]

Okay so there’s a function where A maps to B where S is a subset of B and I’m trying to prove that the inverse compliment of S is the same as the other statement but I’m not sure what the difference is

[hushed pewter]

You are trying to prove that the inverse image of the complement, $f^{-1}(\overline{S})$, is equal to the complement of the inverse image, $\overline{f^{-1}(S)}$.

[twin meteor]

SWR

[marble kettle]

what is the difference between them?

one refers to finding the inverse image of the complement of a set and the other refers to finding the complement of the inverse image of a set

i guess

but maybe i dont understand what that means exactly

[hushed pewter]

The difference is just the wording, hence why we are proving that they are equal

But for starters, $f^{-1}(\overline{S})$ is the set of all $x\in A$ such that $f(x)\in\overline{S}$

[twin meteor]

SWR

[hushed pewter]

And $\overline{f^{-1}(S)}$ is the set of all $x\in A$ such that $x\notin f^{-1}(S)$.

[twin meteor]

SWR

[marble kettle]

okay so S is a subset of B and we are saying that A maps to B so f^-1(S) is all x that exists in A such that f(x) exists in compliment S

[hushed pewter]

okay so S is a subset of B

[marble kettle]

yes

[hushed pewter]

and we are saying that A maps to B
Better to say that f maps A to B

f^-1(S) is all x that exists in A such that f(x) exists in compliment S
Not quite.

Are you writing $f^{-1}(S)$, $\overline{f^{-1}(S)}$, or $f^{-1}(\overline{S})$?

[twin meteor]

SWR

[marble kettle]

the second one

i still dont get how S being a subset of B gives us the information that x is part of A

[hushed pewter]

so f^-1(S) is all x that exists in A such that f(x) exists in compliment S
This is describing the third one. Describing the second one would be like "the complement of all x in A such that f(x) exists in complement S"

[marble kettle]

oh yeah i didnt see the first f^-1(x) i meant this one

[hushed pewter]

Oh. Then yeah, you described it correctly.

[marble kettle]

from there what should i try and work on

[hushed pewter]

What does it mean that f(x) is in the complement of S?

[vocal sleet]

@marble kettle Has your question been resolved?

[vocal sleet]

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[rustic chasm]

hello

[rustic chasm]

can i get help on this problem

On the second part

[graceful surge]

[hushed pewter]

Let $u=t^2$

[twin meteor]

SWR

[rustic chasm]

yup

I got that

[hushed pewter]

!show

[vocal sleet]

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

[graceful surge]

you do the same as the first one

[rustic chasm]

I got lost on the part where you put it in the function

[graceful surge]

it'll just be f(u)

u = t^2, du/2t = du

[rustic chasm]

This is what I got

[graceful surge]

$\int_{a}^{b}g\left(t\right)f\left(t\right)dt\ \ne\int_{a}^{b}g\left(t\right)dt\cdot\int_{a}^{b}f\left(t\right)dt$

[twin meteor]

Combustion

[graceful surge]

should just be $\int_{0}^{64}tf\left(u\right)dt$

[twin meteor]

Combustion

[graceful surge]

and replace the dt

[graceful surge]

yeah keep going

[graceful surge]

t/t = 1

also where did the 2 go?

[rustic chasm]

Ohhhh

I canceled the t but multuplying it with the du/2t to get du/t

But I get what you mean

Wait

U end up with 1/t

Because if you multiple t x 1/2t

= 1/t

[graceful surge]

how

[rustic chasm]

Oh wait

No 1/2

[graceful surge]

yeah

[rustic chasm]

Got it

-25 (1/2)

-12.5 if simplified

[graceful surge]

[rustic chasm]

Thank you very much

I will return if I get stuck on these horrendous ones

[graceful surge]

good luck

[rustic chasm]

Thank you once again

.close

[vocal sleet]

Channel closed

Closed by @rustic chasm

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[vocal sleet]

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[vast shale]

why is my convergence for sqrtx on 0,1 noticeably slower than other integrals

for sin(x):

fs =

' # panels            Integral              Convergence
   2                  6.6044826171e-01      0.0000000000e+00 
   4                  5.6248481067e-01      0.0000000000e+00 
   8                  5.1169091004e-01      9.4758833023e-01 
   16                 4.8584401149e-01      9.7466408978e-01 
   32                 4.7280826737e-01      9.8751823895e-01 
   64                 4.6626233359e-01      9.9380202637e-01 
   128                4.6298235202e-01      9.9691129945e-01 
   256                4.6134060761e-01      9.9845816709e-01 
 '

After:

IntegateTest

fs =

' # panels            Integral              Convergence
   2                  4.5008051550e-01      0.0000000000e+00 
   4                  4.5730093757e-01      0.0000000000e+00 
   8                  4.5909897349e-01      2.0056613293e+00 
   16                 4.5954804321e-01      2.0014104889e+00 
   32                 4.5966028323e-01      2.0003523208e+00 
   64                 4.5968834152e-01      2.0000880614e+00 
   128                4.5969535599e-01      2.0000220141e+00 
   256                4.5969710960e-01      2.0000055039e+00 
 '

for: exp(-(x^2)):

IntegateTest

fs =

' # panels            Integral              Convergence
   2                  7.3137025183e-01      0.0000000000e+00 
   4                  7.4298409780e-01      0.0000000000e+00 
   8                  7.4586561485e-01      2.0109453466e+00 
   16                 7.4658459679e-01      2.0028011114e+00 
   32                 7.4676425465e-01      2.0007034503e+00 
   64                 7.4680916364e-01      2.0001760491e+00 
   128                7.4682039054e-01      2.0000440238e+00 
   256                7.4682319725e-01      2.0000110066e+00 
 '

for x^(5/2):

IntegateTest

fs =

' # panels            Integral              Convergence
   2                  3.3838834765e-01      0.0000000000e+00 
   4                  2.9879149623e-01      0.0000000000e+00 
   8                  2.8897473967e-01      2.0120673730e+00 
   16                 2.8652856790e-01      2.0047207236e+00 
   32                 2.8591777970e-01      2.0017816265e+00 
   64                 2.8576515225e-01      2.0006577467e+00 
   128                2.8572700172e-01      2.0002394568e+00 
   256                2.8571746466e-01      2.0000863798e+00 
 '

for x^(1/2):

IntegateTest

fs =

' # panels            Integral              Convergence
   2                  6.0355339059e-01      0.0000000000e+00 
   4                  6.4328304624e-01      0.0000000000e+00 
   8                  6.5813022162e-01      1.4200277992e+00 
   16                 6.6358119688e-01      1.4456022143e+00 
   32                 6.6555893628e-01      1.4626620297e+00 
   64                 6.6627081138e-01      1.4741562971e+00 
   128                6.6652565730e-01      1.4819988918e+00 
   256                6.6661654898e-01      1.4874051067e+00 
 '

[vast shale]

im guessing it has to do with the infinite slope at 0

[fallow frost]

Ganggangganggang i need hellp like really really quick im stupid and dont get what -2x multiplied with -2 is

And k got a math test

Ion wanna fail

[vast shale]

4x

[fallow frost]

Thanks

You just saved my life

Ily

Im gonna get a decent grade now🙏

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

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[trail seal]

Show that for each value of the parameter m, the line y = mx - (m^2)/4 is tangent to the parabola y = x^2. Find f(m) such that the family of lines y = mx + f(m) has envelope the parabola y = Ax^2 + Bx + C
Someone was helping me earlier but I closed the post and left the q