help-0

[golden elbow]

I'm so bad at math seriously

would it be better to just move into a different area?

it's week 1 of Calculus 2 and this is Calclus 1 review for the first week

[pseudo ice]

And I wouldn't think so, sometimes it's just a lot of work, but of course worth prefacing that with I haven't seen a lot of your work so can't make an informed comment

[golden elbow]

how do you know to simplify like that though?

[pseudo ice]

Well for me it's mostly because that would get me something where I don't need quotient rule, because I find it a bit more annoying to deal with and would prefer product rule where possible

[golden elbow]

ok so always break the fractions

into multiplication

do you tutor?

[pseudo ice]

I do yea, sometimes

[golden elbow]

what do you charge? do you have lots of available time?

[pseudo ice]

I don't know whether I'd say "always" but usually it's a good idea to, some of these it's like follow your nose really

[golden elbow]

i think my nose is plugged 🙂

[pseudo ice]

And I don't have lots of time these days can't say what I charge

[golden elbow]

ok can you help me with question 2

just from what you have seen though, do I have a chance to do this, because I've done 2 -3 weeks of calculus 2 like a couple years ago

and I remember once we started doing trigonometric integrals I quit

am I doomed?

[pseudo ice]

Question 2?

I might reply a bit slowly but can take a look as much as I can

[pseudo ice]

I mean possibly, might be a lot of work for it though but anything is possible

[golden elbow]

3 / srt(t + 1)

so whats the hint for this?

[pseudo ice]

I would say for this one that it might be a tiny bit better to do this by chain rule, rather than quotient rule

[golden elbow]

chain rule so how we would do that

oh right

3 * (t+1)^1/2 ?

[pseudo ice]

-1/2 but yep

[golden elbow]

lol

think I can just learn along the way

or is my math too weak for Calc 2 university?

3 * (t + 1)^(-1/2)

the derivative of 3 is 0

[pseudo ice]

It's possible to learn stuff along the way, though to be fair...

[pseudo ice]

...I'm not sure of the exact content that calc2 covers

[golden elbow]

start with integrals

I sort of got into integrals, but then trigonometric integrals came up

and thats when it gets really complicated

you need to memorize a lot of formulas

[pseudo ice]

It is, but bear in mind it's a constant, you can always "ignore constants" and "pull them out"

[golden elbow]

3 -1/2(t+1)^(1/2)

[pseudo ice]

Yea there can be quite a bit of memorising sometimes that's never been my area, always preferred the stuff you can kinda figure out and stuff

[golden elbow]

wait

3 -1/2 ( t + 1 ) ^ ( -3/2)

(-3/2t - 3/2 )^(-3/2)

[pseudo ice]

This is fine but...

[pseudo ice]

Be really careful with distributing the -3/2, and it's probably better to keep it outside for now

Of course, worth noting that now you have $\frac{-3}2 (t + 1)^{-3/2}$, and you can rewrite that $(t + 1)^{-3/2}$ in multiple steps

[ocean seal]

@pseudo ice

[golden elbow]

can you give me a hint

do you use u substitution

[pseudo ice]

You've already differentiated it

Remember how we had basically rewritten $\frac1{\sqrt{t + 1}} = (t + 1)^{-1/2}$? Kind of the same idea here too...

[ocean seal]

@pseudo ice

[golden elbow]

-3/2 cubert(t + 1 )

why do they have (t + 1 ) sqroot(t + 1)

[pseudo ice]

A very good question

[pseudo ice]

A convincing case that it's not this

Note that $\frac32 = 1 + \frac12$, which might be helpful(!)

[ocean seal]

@pseudo ice

[golden elbow]

(-1 -1/2 ) (t +1)^(-3/2)

[pseudo ice]

While true, maybe use that in the power instead

[golden elbow]

-3/2 (t + 1)^(-1 -1/2)

sorry Im a bit lost

what is the answer?

.close

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[golden elbow]

find the derivative

I got -3/2 (t +1)^(-3/2)

[foggy pecan]

$\text{Note that:}\\a^{-\frac{3}{2}}=\frac{1}{a^{\frac{3}{2}}}=\frac{1}{a^{1}\cdot a^{\frac{1}{2}}}=\frac{1}{a\sqrt{a}}\text{, }\text{ }a>0$

[ocean seal]

Joanna Angel

[lone heart]

@golden elbow Has your question been resolved?

[golden elbow]

why would they want it in that form?

ok can you help me with another

7e^8 / 5s

?

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[random tendon]

There are two batches of products A and B. Lot A has 10 products of type I, and 2 products of type II. Lot B has 16 type I products and 4 type II products. From each batch, randomly select one product. Then, from the 2 products obtained, randomly selected produce a final product. What is the probability that the final product is type I?
The answer was 0.79 and i don't know how it got there.
I have tried multiple ways but the only result I get is 0.8167.

[placid zinc]

Basically a tree diagram

[random tendon]

can you explain more bc i lost there

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@random tendon Has your question been resolved?

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@random tendon Has your question been resolved?

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@random tendon Has your question been resolved?

[shadow orchid]

I have 2 geometry questions

[prime river]

Not able to see them

[shadow orchid]

Oops they never sent- but it’s okay I figured them out! :>

[tulip bone]

close the tick thenn

[lone heart]

@random tendon Has your question been resolved?

[misty mauve]

is the answer 0.775 and not 0.79 by any chance?

@random tendon

[random tendon]

I'm not sure if the answer was 0.79 or not but it was given by my professor. If its possible, can you send your work because it was approximately the answer.

[lament forge]

i also got 0.8167
i have no idea how you'd get 0.775 or 0.79

[prime badge]

i get 0.775

,calc (1/2)(10/12)(4/20) + (1/2)(2/12)(16/20) + (10/12)(16/20)

[ocean seal]

Result:

0.81666666666667

[prime badge]

intresting

,calc (1/2)(10/12)(4/20) + (1/2)(2/12)(16/20) + (10/12)(15/20)

[ocean seal]

Result:

0.775

[prime badge]

that's how i got it, there's a typo in 15/20

[random tendon]

I think you were right, maybe the result is 0.816... it was approximately the answer, i think my professor has some mistakes

Thank you all for answering my question 😭

I appreciate it

[cosmic hazel]

If sectheta + tantheta = 1/k then show that coesectheta = 1+k^2/1-k^2

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[lone heart]

@vivid silo Has your question been resolved?

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[ruby elbow]

how do I use the Pythagorean theorem

in this question

it asks me to figure out the hypotenuse of triangle ABD

all I know is that angle ABD is a right angle tho

i know the theorm is A^2 + B^2 = C^2

but how would I apply this to triangle ABD

A^2 + B^2 = D^2???

wait

I know that AB is 4 units long

and I know that DB is 3

ahhh

4^2 + 3^2 = hypotenuse

make sense

25 = hypotenuse?

is 25 the hypotenuse of ABD?

[languid hinge]

the sqrt of 25

its a special triangle

[ruby elbow]

?

[languid hinge]

345

[ruby elbow]

wha

now I am confused

4^2 = 3^2 = C^2

25 = C^2

waht

[languid hinge]

yes, its correct

take the square root of both sides

so sqrt(25)=sqrt(C^2)

sqrt(25) = C

[ruby elbow]

what do you mean square root tho

im confused

[languid hinge]

do you know the pythagorean theorem or is it your first time working with it?

[ruby elbow]

i know it

i used it in my math homework hundreds of times last year probably lol

[languid hinge]

so a^2+b^2=c^2

we have our triangle ABD

AB = AC/2

[ruby elbow]

wym

AB = AC/2

ABD is equal to CBD we are just trying to prove it

AB and CB are both equal to 4

DB is 3

AB = 4

DB = 3

[languid hinge]

the question is asking for the pythagorean theorem

[ruby elbow]

4^2 + 3^2 = C^2

25 = C^2

[languid hinge]

that's right

so what times itself results in 25?

[ruby elbow]

5 *5

[languid hinge]

yes, so c=5

[ruby elbow]

yea

[languid hinge]

thats what the question is asking for, right?

[lone heart]

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[lone heart]

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[alpine sable]

was maths invented or discovered

[dense iris]

How to integrate this??

[sour verge]

This is a deeply philosophical question that doesn't have a straight-forward answer.

It goes back quite far away, even in Ancient Greece, with opposing views proposed by Aristotle and Plato.
You can think of Aristotle's formalism as the idea that we set some basic truths called axioms, and then build/create new truths that logically follow from them. In this viewpoint, mathematical truths are invented.

Plato had this concept of a separate world where ideas of things live (for instance, we mathematically can comprehend the "idea" of a circle, but in real life there is no such thing as a circle due to imperfections in all around us). In such a conception of reality then, the "ideas" of mathematics have always existed in this world of ideas, and are simply uncovered as we progress.

You could do some more research on the subject to get a better grasp, but in short it's a matter of belief on a fundamental level, i.e. there's good arguments to justify why one would believe any of those two points of view.

[alpine sable]

whats ur personal view on this question

[foggy pecan]

mathematics does not need people for its existence, but without people it will be mute, no one will talk about it, that is, it exists but it does not exist, this is naturally a human view, and if there is something other than humans that can deal with mathematics, then people are not needed for mathematics

hence, maths is being discovered

[alpine sable]

oooohh like language?

thats very interesting point

[sour verge]

That's one view yes, the platonist one.

[sour verge]

I like to think these truths in mathematics are rather synthetic inventions of ours to make sense of things and put order in what is usually quite unordered like the world.

But then again, this question is more philosophical than mathematical in nature, and it's up to you to forge your opinion of it, as there is not a "right" answer.

[alpine sable]

alr

thanks everyone for answering my question

.close

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[alpine sable]

.reopen

[lone heart]

✅

[alpine sable]

r u typing smth

[sour verge]

No it's alright I was just commenting on the language argument but it's not necessary and I have pasta to boil 🙂

[alpine sable]

ohh alr gl with ur pasta

[sour verge]

Thanks!

.close

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[ocean seal]

QuasiStar 超新星

[livid sage]

notice how (y - x)^2 = (x - y)^2

:O

[ocean seal]

QuasiStar 超新星

[livid sage]

[ocean seal]

QuasiStar 超新星

[livid sage]

i suppose you could

although if we were to substitute, i probably wouldn't include the cubes

ah alright

it most certainly is not but i suppose you could solve it however you want

[ocean seal]

QuasiStar 超新星

[livid sage]

🤔

that would be correct

i do not believe that would help you here

(if you are wondering...
from this point, this problem is immediately broken with the ||existence of a somewhat known identity involving the sum of 3 cubes||)

left in spoilers in case that wasn't necessary

it's pretty hard to derive so you might need to look it up

this wasn't what i was thinking of, but it is true can use this to solve your problem

try seeing what happens when you re-substitute a = x - y ..etc into that expression

[ocean seal]

QuasiStar 超新星

[livid sage]

oh

when i said "re-substitute" i meant replace a with x - y and so on

it will become much more clear why this is necessary once you see what happens

[livid sage]

but yes this is very nice looking

no i mean the other way around

like instead of having things in terms of a, b, and c... have them in x, y, and z

[lone heart]

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[unreal flame]

how many unique combinations of three colours can you make given 4 unique colours?

[unreal flame]

as in, red green blue, red green green, ect

[naive valley]

so like red red red is a valid combination?

[unreal flame]

yes

also, in extension, how many unique combinations excluding isomorphs?

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[lone heart]

Please don't occupy multiple help channels.

@young patrol Has your question been resolved?

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[young patrol]

I am having trouble understanding difference quotients, in the sense of radicals especially

[lone heart]

Please don't occupy multiple help channels.

[tulip bone]

wdym

[proud creek]

Just Rationalize with $\sqrt{2x + 17} + \sqrt{2x + 5}$

[ocean seal]

Gowtham

[young patrol]

Doing this problem is an example of difference quotient

[tulip bone]

mhm

[young patrol]

I’m p sure

[tulip bone]

yeah, that sounds like a good idea

[young patrol]

What is the value after

Or is that all you can do

[proud creek]

wait i done that

\dfrac{2}{\sqrt{2x + 17} + \sqrt{2x + 5}}

dfrac doesnt work?

\frac{2}{\sqrt{2x + 17} + \sqrt{2x + 5}}

[young patrol]

Ok we still have a denom of 6

Since you are rationalizing that wouldn’t you multiply

[young patrol]

This by 6 as well

[proud creek]

$\frac{2}{\sqrt{2x + 17} + \sqrt{2x + 5}}$

[ocean seal]

Gowtham

[proud creek]

this is the ans i got

on further rationilizing we get the queston

[young patrol]

Is this the correct way of setting up

[proud creek]

use $(a + b)(a - b) = a^{2} - b^{2}$

[ocean seal]

Gowtham

[proud creek]

for rationilizing

[young patrol]

So that is correct (wrote it other way around)

Square root 2x+17 times 2x+17 is just

2x+17

Ughhh

I’m gonna write it out

I wrote it clearer looking for my mistake @proud creek

This simplified to me is 2x+17-2x+5

(A+B) in this case is the opposite already of f(x+6) - f(x)

😢

[unborn monolith]

what the question you need help with? is this it?

[young patrol]

Yes!

[proud creek]

but, after rationalizing it has roots in denominator

[unborn monolith]

write it as: $\newline \frac{\sqrt{2(x+6)+5}-\sqrt{2x+5}}{6}$

[proud creek]

the final ans comin' is

[ocean seal]

TheLord26

[proud creek]

$\dfrac{2}{\sqrt{2x + 17} + \sqrt{2x + 5}}$

[ocean seal]

Gowtham

[proud creek]

As said in the question, Numerator A has no surds or radicals

@young patrol

[unborn monolith]

multiply top and bottom by $\sqrt{2(x+6)+5}$ and $\sqrt{2x+5}$

[ocean seal]

TheLord26

[unborn monolith]

anyways im going, ive got other things to do, sorry

[young patrol]

Thanks!!

[young patrol]

In here, we’d do 6(2x+17+2x+5) for the denominator right?

[young patrol]

.

[young patrol]

Like where does the 6 go in this

I’ll ask again tomorrow I guess

.close

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[torn isle]

Can anyone help me for a?

Not sure how to form the differential equations

[placid zinc]

Rate of change of x1
= amount of salt entering tank 1
- amount of salt leaving tank 1

The amount of salt moving depends on the concentration of salt

[lone heart]

@torn isle Has your question been resolved?

[torn isle]

Yea, I know this

But actually I’m stuck at the beginning

Cause initially there’s volume of water in the tank, one is brine, one is pure, not sure how to form

[placid zinc]

How to form which?

That's the solution to a

[torn isle]

x’_1(t)

Like initially tank 1 contained brine

And tank 2 contain pure water

I’m not sure how pure water difference with brine

[torn isle]

This

[placid zinc]

Seems irrelevant to part a

[torn isle]

Huh? Part a asked us to find differential equations for tank 1 and tank 2 right?

[torn isle]

Then I form this, but stuck for the input part

[placid zinc]

Your picture contains the differential equations for tank 1 and tank 2.

[torn isle]

Cause I asked in an app, then they replied this answer, which I think the answers are wrong

[torn isle]

Yea, and I wanna know how to form the differential equation

I need to form x_1’ and x_2’

[placid zinc]

Is a part of the picture not making sense?

[torn isle]

?

[placid zinc]

How to form the differential equation is in the first picture you sent

[torn isle]

Yea, that’s what I’m asking, I want to form 2 equations

So that I can proceed for the following steps

[placid zinc]

Thank god you sent me a picture detailing how that's done!

[torn isle]

?

[placid zinc]

Do you want an explanation of any part of this picture?

[torn isle]

I don’t think the answers provided here is correct

[placid zinc]

OH

Okay I think you are correct

[torn isle]

The questions didn’t mention any about 25 and 50, so how can they use that to solve

[torn isle]

That’s why I’m asking, from my own workings, how do I put the input for both tank

I think for output is correct, just input

Because initially for tank 1, it’s brine, and tank 2 for pure water

[torn isle]

I just don’t know how to form for input

[placid zinc]

Note their initial values don't matter for part a

However the concentration into tank 1 isn't given by "2"

Concentration into tank 1
= Concentration of tank 2
= x2(t) / 1

Huh. That's what they put for the concentration out I think

[torn isle]

I think I refer to textbook before, it’s not solved like this

And that’s why my output is like that

Are you taking about input?

[placid zinc]

input?

[torn isle]

Erm, nvm, maybe I should ask in calculus channel

[placid zinc]

I am talking about the concentration of fluid going into tank 1

[torn isle]

Anyway, thanks for the help

[torn isle]

I don’t get what you meant, I just wanna know how to form the input and output for each tank

[placid zinc]

Concentration into tank 1
= Concentration of tank 2
= x2(t) / 1

That's the input into tank 1.

Well, that, when multipled by the flow rate.

[torn isle]

Then input of tank 2 is x1(t)/1?

[placid zinc]

You know what? Go ask. Cya!

[torn isle]

.close

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[gloomy verge]

which quadrants do all the inverse trig functions belong to (arcsin, arctan, arccos, arcsec, arccsc, arccot)

[coarse vale]

depends from function to function I think

Like for sin^(-1) it's [-π/2,π/2] which is 1st and 4th quadrant.
For cos^(-1) it's [0,π] which is 1st and 2nd quadrant.

[gloomy verge]

how do i read that notation

like

u said -pi/2 was 1st quadrant

but then u also said

0

was the 1st quadrant

[buoyant saddle]

it’s the y axis

but below zero

it’s an angle

it makes up Q4

[coarse vale]

make a unit circle, imagin a line making angle with +ve x axis, then +ve x axis is 0 and the +ve y axis is π/2 and the circle continues till u reach +ve x axis again in which case it's 2π

[buoyant saddle]

arcsin is Q4 to Q1

[foggy pecan]

look:

[buoyant saddle]

essentially the right side of the graph

yea exactly

[gloomy verge]

ohh

ty

thats clutch

[buoyant saddle]

the domain of the inverse trig functions is the range of the trig functions

[gloomy verge]

thats good 2 know

[buoyant saddle]

and vice vera

[gloomy verge]

ty

.close

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[alpine sable]

is the equation gonna be

$x^{2} + y^{2} + z^{2} = r^{2}$

[ocean seal]

Gary

[foggy pecan]

yes

correct

[rocky imp]

ya

[alpine sable]

in the back of the book idk why the

x^2 + y^2 + z^2 = 1

but not r

hmm

[foggy pecan]

r = 1 in thier case

[rocky imp]

ig they took radius = 1 unit

[alpine sable]

oh

i see

so its correct

[foggy pecan]

ofc

[alpine sable]

thank u guys

!!

[foggy pecan]

yw!

[alpine sable]

.close

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[fair osprey]

close

you missed a term

[vale wigeon]

not close at all you can't just erase the roots

[fair osprey]

i think he just missed the √b in rationalizing?

[vale wigeon]

might have

no work shown tho!

[fair osprey]

true but he had the same problem a bit earlier so i just assumed

[kind sparrow]

You also gave them the answer and then they closed it. If anything this is showing they have learned nothing from that.

[fair osprey]

i thoguht they figured it out? didn't realist that i did

[vale wigeon]

!nosols

[lone heart]

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.

@noble glen Has your question been resolved?

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[hazy juniper]

(Ignore everything I wrote except the question itself)
What does this even ask??? To remove a corner from a cube to form a figure means what? And why does removing a corner leave a leftover portion EB? The answer key lists answer to A as PE ≈ 30 cm, PR ≈ 33.9cm, B as ≈69⁰

[blazing oriole]

i can your calculations and you are going right

just the calculations are a bit wrong

[torn lotus]

Please check

[blazing oriole]

PA = 24 cm

EB = 6 cm

so PA' = 24 - 6 = 18 cm

Wait, lemme show you the diagram

[hazy juniper]

kk

[ornate ginkgo]

Idk if it will be a triangular pyramid exactly but you can calculate the sides with pythagorean theorem

[hazy juniper]

i just don't fundamentally understand the question

how do you remove a corner from a cube and what exactly am i finding like what even is this shaded aera

[ornate ginkgo]

Cosider cube to be a solid shape

Wait let me draw it out

[torn lotus]

Help

[ornate ginkgo]

Open a new help channel this is already occupied

[blazing oriole]

this is how it should look

[hazy juniper]

So that's whay of a cube of 24 cm means?

[blazing oriole]

now take triangle PHE and find PE

and then if you take triangle PSR with S as the right angle and you will get PR

[blazing oriole]

it means the side of the cube is 24 cm

[hazy juniper]

Ahh okay the question makes a lot more sense now

[blazing oriole]

and regarding how do you remove a corner from a cube

assume it to be a solid
not a hollow cube

so you can just cut one corner of it

dats how u can imagine

[hazy juniper]

Okay, I understand what's going on now

Thanks!

.close

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[blazing oriole]

np!

this is how it would look btw

[ornate ginkgo]

Nvm

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[oblique plinth]

Can someone help me through this. im kinda lost in how it turns into 1/2 * r

[boreal lily]

you just factor with with 1/2r (cos(theta)^2 +sin(theta)^2) = 1/2r

@oblique plinth

[oblique plinth]

yeah im sorry. i got it wrong thinking that the cos(0) -rsin(0) were in the same bracket.

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[silent idol]

I'm getting my answer as 1/ root 3

idk if that's correct

[vale wigeon]

show work?

[silent idol]

alr gimme a sec

[raw stream]

preboard paper?

[silent idol]

dont mind d bad writing

[raw stream]

What is sin/cos?

[silent idol]

yeah

[raw stream]

Which school?

[silent idol]

1/root 3

[raw stream]

nono

How can it be represented

[silent idol]

what?

[raw stream]

ok tan x = sinx/cosx

[silent idol]

tan 30?

tan 30 = 1/root 3

[raw stream]

yes

[silent idol]

sinx/ cosx = 1/root 3

what is the answer tho?

1/root 3

or tan 30?

[raw stream]

do one thing write cot as cos/sin, and pull out the sin from the brackets but don't open the bracket it will help for cancellation

[raw stream]

yes

[silent idol]

thats what i did

[silent idol]

.

[raw stream]

Oh i didnt see sorry

[vale wigeon]

ok yeah seems to check out

you simplified it all down to tan(θ)

[raw stream]

Yes

[silent idol]

alr

[raw stream]

Answer is correct tho

[silent idol]

ty ❤️

.close

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[raw stream]

Which school?

[silent idol]

dps

[raw stream]

where?

[silent idol]

dms

[lone heart]

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[fast zephyr]

Can someone explain what the limit lim in the context of derivatives means?

[vale wigeon]

do you know what a limit is outside the context of derivatives

[fast zephyr]

to be honest, not really

[vale wigeon]

ok well... think you might want to learn that before derivatives

[fast zephyr]

ok

any tips where i can learn it?

[vale wigeon]

or look at shit like khanacademy. maybe also professor leonard's vids on youtube

ochem tutor might be good as well

[fast zephyr]

ok

thx

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Channel closed due to the original message being deleted.
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as this action is irreversible, and this channel may abruptly lock.

[median thicket]

What I understand is that the amplitude will be drawn vertically as the 5 Pa mark among many other marks and the period will be drawn as the 0.03 sec mark among many other marks horizontally. This is my question… For the graph, how many marks will l need vertically and horizontally and how will those number values be spaced out? I’m assuming I need to draw mine similarly to some of these examples:

[vale wigeon]

!msgdel

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The original post of this help channel has been deleted, and it will abruptly close and possibly lock. (This is irreversible.) Please claim a new channel, and don't delete the first message of any future channel you claim.

[median thicket]

.close

[median thicket]

Sorry did I do something wrong? Do I need to reset this channel?

[vale wigeon]

you need to open a new one

[quasi vector]

you should open a new channel

[vale wigeon]

you did do something wrong on a technical level

[quasi vector]

and not delete the first message you send there

[median thicket]

Ok

[vale wigeon]

you deleted your first msg

you're not supposed to do that

[median thicket]

Oh whoops. Sorry

.close

Oh it won't let me close it

.close

[median thicket]

it won't let me close now. Sorry, i didn't realize

[vale wigeon]

it's already closed

it'll release in 10 minutes

just go open a new one

[median thicket]

Oh thank you

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[dense iris]

Integration

[dense iris]

I know when solving for dz I will be. Taking x,y constant but then will the new equation ??

[lone heart]

@dense iris Has your question been resolved?

[flat roost]

what's your question exactly?

[open compass]

bc everything else is constant u can just use u sub here

if ur asking what the antiderivative evaluates to

[dense iris]

You have to evaluate the above double Integral , so can anyone provide the solution... Bc I keep getting it wrong

[flat roost]

lemme try

we dz then dy then dx right

nvm this is too confusing sorry

[dense iris]

Yup

[open compass]

i got -ln2 + 3/4

it should be possible if u use u substitution

it does require you to know the integral of lnx though

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@dense iris Has your question been resolved?

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[waxen citrus]

Sorry forgot how to do these questions… Find the equation of the tangent to the curve y = x^3 - 3x^2 + 2 at the point on the curve where x = -1.

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@waxen citrus Has your question been resolved?

[dense iris]

Yup the answer is write

Thanks

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[wet vector]

need some help plz

[sleek girder]

try to phrase the question in terms of a variable

[wet vector]

ok i have an answer

i used a solution from google to help me

[sleek girder]

just to be sure, if you read carefully, here should be the things you should take note:

1. advertised dress price is 100%;
2. current dress price is 20% lower than advertised dress price;
3. current dress price is 104% cost price
4. advertised dress price is ?% cost price

[wet vector]

but i dont get one thing

why do they make the two expressions equal ?

[sleek girder]

they should be both equal to the advertised price

[sleek girder]

1. advertised dress price = 100% P
2. current dress price = 80% P (= R)
3. current dress price = 104% C
4. advertised dress price = ?% C

P = advertised dress price
R = current dress price
C = cost price

so, per clues 2 and 3, 80% P (= R) = 104% C

any other questions, @wet vector?

[wet vector]

makes sense

got it now

thank u

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[rigid dock]

why is this false?

[vale wigeon]

,w x^2+3x+9=0

[ocean seal]

Results provided by WolframAlpha

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

Requested by ann_dec

[cinder tundra]

Who says?

[rigid dock]

my solution

[cinder tundra]

Can you show it the picture of the solution?

[rigid dock]

It's a multiple choice question so I just have the answers and no explanations...

[vale wigeon]

that's strange

it would appear true

[cinder tundra]

Yes, it must be a typo or a misreading by all us

More likely a typo

[rigid dock]

alright I will check this

thank you

.close

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[vocal tapir]

this is true, right?

@vale wigeon Нашият лектор просто извади e на една задача

и направо ми се прищя да се гръмна

а не e^n

[minor bone]

ok im going to sleep

[vocal tapir]

и си викам как по дяволите

[vale wigeon]

💀

[vocal tapir]

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[cerulean junco]

why tr(α'α) =1?

[cerulean junco]

it is a rectangular matrix how the hell are they able to find trace of it ?

[worn fox]

' is transpose?

[cerulean junco]

yes

and tr() is trace .

[worn fox]

If α is an mxn matrix what size is α'α?

[dense iris]

Can you please share the solution

[cerulean junco]

mxm? oohhhhh

[worn fox]

Nearly, nxn

Point is it'll always be square

The other order αα' aswell

[cerulean junco]

okay ,but why 1 ?

[worn fox]

What is the size of α' if α is nx1

[cerulean junco]

nxn !

[worn fox]

Of α'?

[cerulean junco]

1xn !

[worn fox]

Yes

So size of α'α?

[cerulean junco]

1x1 ... oh okay.fair enough .

[worn fox]

Yeah so α'α is a number, they're telling you it's the number 1

[cerulean junco]

thanks

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[valid verge]

Hi, im stuck on how i would find the coordinates of the maximum point

[pseudo ice]

I think it's safe to assume you know the max value is $\sqrt{13}$ - you want to solve $\sqrt{13} \sin(x + 0.98) = \sqrt{13}$ with the solution $x$ between 0 and $\pi$

[ocean seal]

@pseudo ice

[valid verge]

would i be subbing in root 13 as the x and then get y with that?

[pseudo ice]

Other way around: sqrt{13} is the y, you want the x

[valid verge]

ohhhh

oh would i set y equal to root 13 into 3sinx+2cosx equation?

[pseudo ice]

Just put it into the changed form, the new one you found, as that makes your life easier

You're then reduced to solving $\sin(x + 0.98) = 1$ with $0 \leq x \leq \pi$ of course

[valid verge]

tried that to find x but then got math error in calculator

[ocean seal]

@pseudo ice

[valid verge]

Oh

[pseudo ice]

How did you do it?

[valid verge]

Moved everything to make x subject

[pseudo ice]

Should have left it as sqrt{13} rather than rounding it to 3.61 I think is why

[valid verge]

oh yeah that was the reason

answer is (89.02, root 13) for coordinates?

[pseudo ice]

You want the solution in radians but I think you have the idea

Instead you want to work out pi/2 - 0.98 for the x coordinate but that's pretty much it

[valid verge]

i dont understand where the pi/2-0.98 comes from would you mind explaining it

[pseudo ice]

Replace 90 degrees with pi/2 radians

[valid verge]

oh i see thank you

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[coarse fiber]

Let M ⊆ R be a subset of real numbers. We say that h is an accumulation point of M if, for all error bounds ε > 0, there exists an element x ∈ M \ {h} such that |h - x| < ε.

(a) Show that if x₀ ∈ R is an upper bound of M and an accumulation point of M, then x₀ = sup(M).

(b) Show or refute the following statement: If x₀ = sup(M), then x₀ is both an upper bound and an accumulation point of M.

Note that, analogous to the statement in (a), it also holds: If x₀ is a lower bound and an accumulation point of M, then x₀ = inf(M). (Not to be proven but nice to know).

I have no idea what I need to do here and my brain is slime right now

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@coarse fiber Has your question been resolved?

[coarse fiber]

@craggy dagger

[lone heart]

@coarse fiber Has your question been resolved?

[lone heart]

@coarse fiber Has your question been resolved?

[alpine sable]

assume there exists a smaller upper bound y < x₀,since x₀ is an accumulation point, for any ε = (x₀ - y)/2 > 0, there exists x ∈ M \ {x₀} such that |x₀ - x| < ε,this implies x > y, contradicting the assumption that y is an upper bound ig so x₀ is the least upper bound, or supremum, of M, as no number smaller than x₀ can be an upper bound.

Ich hoffe "es hilft dir".

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@coarse fiber Has your question been resolved?

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[raven flame]

f

[lone heart]

Please don't occupy multiple help channels.

[raven flame]

hellp

i need help with this question

[tacit arch]

close one of your channels

[raven flame]

ok

[lone heart]

@raven flame Has your question been resolved?

[raven flame]

help!

[lone heart]

@raven flame Has your question been resolved?

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[mental breach]

may I know how do i solve this

[mental breach]

I was thinking of e^(-0.2(0)) - e^(-0.2(t)) which gives 1-e^(-0.2t), but i dont think its right

[tacit arch]

what's Z?

doesn't look like the integrand depends on Z at all

[mental breach]

should be Z(s)

[tacit arch]

...

what's the whole context

[mental breach]

Z(t) is a standard brownian motion, i assume it has nothing to do here

it's part of a question, i'm just stuck midway differentiating this

[tacit arch]

well use exponent rules to factor the stuff with t out of the integral

then product rule and ftc from there

[mental breach]

I dont quite understand :c

[lone heart]

@mental breach Has your question been resolved?

[lone heart]

@mental breach Has your question been resolved?

[tacit arch]

,tex .exp rules

[ocean seal]

riemann

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[mystic tapir]

I have no idea where to start. May someone help me?

[finite flax]

you are given displacement as a function of time.
velocity as a function of time is the derivative of displacement

[mystic tapir]

so would I plug the given time into the equation of motion?

[finite flax]

you can just use

$\frac{s(t_2) - s(t_1)}{t_2 - t_1}$

[ocean seal]

Melvin Eugene Punymier

[finite flax]

which is the average velocity over an interval, where s is the (signed) displacement function

[mystic tapir]

okay thank you, I can work on it from here :)

.close

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[placid hemlock]

Would e just be counted only once in any instance of S_k? Tbh I'm not totally sure what the bottom sum means

[lone heart]

@placid hemlock Has your question been resolved?

[placid hemlock]

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[old sluice]

Yo can someone please help me with this equation

[old sluice]

I multiplied -3 by 5x-2y equals -10 but I’m stuck because there is still nothing to cancel

[languid hinge]

you can isolate "y" in the first equation, then substitute the "y" with it's value

5x-2y=-10
-2y=-5x-10
2y=5x+10

[old sluice]

Wait nvm I’m so dumb it didn’t work because the -3 I was multiplying by was a negative instead of a positive 🤦‍♂️

[languid hinge]

you can then change the "6y" in the second equation to "15x+30"

3x+15x+30=66

[old sluice]

This is the way my teacher taught me how to do it

I think I’m doing it right so far

Do you think so?

[languid hinge]

if the teacher taught you to do this way, i think it's better for you to stick with it

it's right

[mighty crest]

its kinda annoying just isolate everything to make y=ax+b

and then sub

u get the answers

but if the teacher wants to use that method

go for it

[old sluice]

The method I’m using is kind of annoying

I’ll send you another problem can you show me your method

[mighty crest]

sure

[old sluice]

I actually have no idea how to do this one I can’t make the y or x the same number with a different sign

[mighty crest]

do you understand this

i transformed the first into y=1+2x

[old sluice]

This is the only part I understand

How did you turn it into this

[mighty crest]

ah ok wait

this

to get rid of the - sign

minus sign

[languid hinge]

he just inverted the "-" sign to make calculations easier

[mighty crest]

yeah

[languid hinge]

so positive becomes negative and negative becomes positive

[mighty crest]

parentheses because you need to divide each term

[old sluice]

Ngl this way seems way more complicated

[mighty crest]

not just one

[old sluice]

Maybe it’s bc I’m just learning it but idk

[mighty crest]

really ok then forget this

yes later on maybe u will use it

i find it much faster

especially if u got a calc

[old sluice]

Could you teach me how to do that problem in this form pls

[languid hinge]

to make the calculations way more easier, we can make it have only one variable instead of two. this is called the "Substitution Method":

5x - 2y = -10
-2y = -5x - 10
2y = 5x + 10 (inverted the signs)

with our new values for "y", we can calculate "x" in the next equation

3x + 6y = 66
3x + 3(5x + 10) = 66 (changed 6y to 3 times 2y)
3x + 15x + 30 = 66
18x + 30 = 66
18x = 36
x = 2

now that we know "x", the next step is to find "y"

3x + 6y = 66
6 + 6y = 66 (calculations with the x value)
6y = 60
y = 10

[old sluice]

Ik how to do substitution method it was easier but my teacher wants us to only do elemination

[languid hinge]

if that's what your teacher is asking for, i think it's better for you to do it by elimination then

it's working anyway

[old sluice]

Alr

Can you show me what I did wrong here

I searched it and it’s wrong 🤦‍♂️

[languid hinge]

you didnt change the result of the first equation

it needs to be -18

[old sluice]

I see what I did 🤦‍♂️

Don’t you mean -27?

3x-9

[languid hinge]

you multiplied the first equation by two, right?

[old sluice]

No by 3

[languid hinge]

then it should be 9x+6y=-27

[old sluice]

Ok thanks that’s what I got

Thanks for your help ❤️

.close

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[vapid shuttle]

I'm trying to think of the contrapositive statement for this

is it

suppose v1, .... vj do not form a basis of U, then => v1,.... vn do not form a basis of V?

or

suppose v1, .... vj do not form a basis of U, then U is not a subspace of V?

So you're saying I should be looking for a counter-example instead of attempting to prove it by contrapositive?

Charade

I was more wondering about

my contrapositive statement

not the actual question itself

Is either one of them formed correctly?

What?

If the contrapositive statement is true then the original statement is true

and you just said the original statement is not true and were giving a counterexample

????

[keen plinth]

what the heck

just happened

[vapid shuttle]

I think they were using chatgpt

..

[keen plinth]

bruh moment

[rose sigil]

[vapid shuttle]

Could you uh

maybe respond to my actual questio

since they wouldnt

XD

Okay I'm going to restate my question since it got a bit confusing above

"I'm trying to think of the contrapositive statement for this
is it
suppose v1, .... vj do not form a basis of U, then => v1,.... vn do not form a basis of V?
or
suppose v1, .... vj do not form a basis of U, then => U is not a subspace of V?"

I'm not saying the statement is provable, I'm just trying to problem solve and see if this might work

[worn fox]

i would just do a counter example, pick V = R^4 or smth

[vapid shuttle]

okay

V=R^4

[keen plinth]

the second one doesnt seem right

because if you're saying v1, ..., vj don't form a basis of U

you're already assuming U is a vector subspace

[vapid shuttle]

so the contrapositive statement would be the first one

[vapid shuttle]

basis vectors (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)

[keen plinth]

well the first one also doesnt seem completely right

[vapid shuttle]

take a subspace the xyz ( is it called a hyperplane? )

then

[keen plinth]

since the original statement is (V is a vector space and v1, ..., vn is a basis and U is a subspace and v1, ..., vj is in U and vj+1, ... vn is not in U) => v1, ..., vj is a basis of U

so the contrapositive would be v1, ..., vj is not a basis of U => negate this whole thing (V is a vector space and v1, ..., vn is a basis and U is a subspace and v1, ..., vj is in U and vj+1, ... vn is not in U)

[vapid shuttle]

is the trick that technically (1,0,0,0) say as a basis vector of R4 isn't in any 3D subspace ?

[night geyser]

no

(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0) is the basis for a three-dimensional subspace of ℝ⁴

[worn fox]

you're looking to construct a space that say has e1 and e2 in it, does not have e3 or e4 in it, but e1&e2 is not a basis

[night geyser]

in fact im confused why people are saying to find a counterexample

[vapid shuttle]

I thought that it would be provable

initially

but I haven't thought of a proof

(nor a counterexample)

[vapid shuttle]

suppose v1, ... .vj is not a basis of U => V is not a vector space or v1, .... vn is not a basis of V or U is not a subspace or v1, .... vj is not in U or vj+1, .... vn is in U ?

I think I missed u r point

[keen plinth]

it would be or

[vapid shuttle]

instead of ands

I see

[keen plinth]

yeah

[night geyser]

i would not prove this by contrapositive

my hint is linear independence

are you familiar with the theorem that says, given a vector space of dimension d, a linearly independent set of size d is a basis?

(where the elements of the set are from your vector space)

[worn fox]

? this isnt true

[vapid shuttle]

I am familiar with that theorem from my old linear algebra class but not with abstract vector spaces like in this class

like for F^(n) I know that

[vapid shuttle]

linear algebra fight

Okay I'm not sure what to do because some are giving me advice to try a counterexample and some are giving me advice to prove it

Im willing to go either way

[keen plinth]

find a counterexample

[worn fox]

do this for R^4

[keen plinth]

the statement isnt true

[vapid shuttle]

okay

I have an idea

[night geyser]

am i being gaslit

how is it not true

[vapid shuttle]

😭

[keen plinth]

lmfao

[worn fox]

try doing this lmao

[foggy pecan]

all bases in space are equinumerous, and every system of linearly independent vectors that is equinumerous to a basis is also a basis

[vapid shuttle]

(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)

Take a subspace that has
(1,0,0,0) and (0,1,0,0) in it, but does not have the other two.
Then this subspace clearly contains the xy-plane.

But we want to say that e1 and e2 do not span the xy-plane?

[night geyser]

icaird i am very interested in your counterexample

[worn fox]

the question is just saying U is atleast j dimensional and doesnt contain some finite number of vectors, nothing stopping it being higher dimensional if your V has enough wiggle room

[night geyser]

feel free to dm it to me

[keen plinth]

isnt there a counterexample in R^3 already

[worn fox]

probably

[vapid shuttle]

I might be more interested than u

[worn fox]

i was worried about trivialness, i.e. j = 1 or j = n

[keen plinth]

||e3 is in {x+y = 0}, but e1, e2 are not||

[vapid shuttle]

?

[night geyser]

sorry i was confused by icaird's objection, you are indeed looking for a counterexample

[worn fox]

weird way to order the std basis of R^3

[night geyser]

i thought icaird was objecting to this line

this line is useful for figuring out what youre looking for for a counterexample

[vapid shuttle]

r u saying to take the subspace {x+y=0}
and that this subspace has e3, (0,0,1)
because 0+0=1.
And then but it does not have (0,1,0) or (1,0,0).
But e3 does not span it?
because there are vectors like (1/2, 1/2, 0) that e3 cannot reach

I think I get that

[night geyser]

lets take the vectors (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)

can you find a set of, let's say, three linearly independent vectors that cannot construct 3 of the vectors in this set?

[vapid shuttle]

I'm a bit confused wym, but you can take the first 3 (linearly independent) and they cannot construct the fourth vector

[night geyser]

but they can construct (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0) [trivially]

which is 3 of the vectors in this set

[vapid shuttle]

yes

ZAC

Was there a point to that

that I missed

why the stare

[worn fox]

idk i was just staring at you bc nami abandoned you

[vapid shuttle]

that's alright

I think I solved it anyways

ty for the help again

.close

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[foggy garden]

So this is one way to solve it (question 3) and the answer is correct but surely there would be a simpler solution:
\left(\frac{5\sqrt{5}+5\sqrt{7}+10}{2}\right)\left(\frac{5\sqrt{5}+5\sqrt{7}+10}{2}-10\right)\left(\frac{5\sqrt{5}+5\sqrt{7}+10}{2}-5\sqrt{5}\right)\left(\frac{5\sqrt{5}+5\sqrt{7}+10}{2}-5\sqrt{7}\right)

[foggy garden]

$\left(\frac{5\sqrt{5}+5\sqrt{7}+10}{2}\right)\left(\frac{5\sqrt{5}+5\sqrt{7}+10}{2}-10\right)\left(\frac{5\sqrt{5}+5\sqrt{7}+10}{2}-5\sqrt{5}\right)\left(\frac{5\sqrt{5}+5\sqrt{7}+10}{2}-5\sqrt{7}\right)$

[ocean seal]

:)

[foggy garden]

ahh sqrt of that mb

$\sqrt{\left(\frac{5\sqrt{5}+5\sqrt{7}+10}{2}\right)\left(\frac{5\sqrt{5}+5\sqrt{7}+10}{2}-10\right)\left(\frac{5\sqrt{5}+5\sqrt{7}+10}{2}-5\sqrt{5}\right)\left(\frac{5\sqrt{5}+5\sqrt{7}+10}{2}-5\sqrt{7}\right)}$

[ocean seal]

:)

[lone heart]

@foggy garden Has your question been resolved?

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[cinder sundial]

Why it is theta

The angle that Mickey is pointing at

[timid lark]

Why are both rays labeled "V" - that's annoying

The "V" on the left is tangent to the circle at the point where the "top" side of the triangle in the circle intersects the circle

So the "top" side and the tangent V form a right triangle

The right V extending from the top side of the triangle is parallel to the right V (which is tangent to the circle at the bottom side of the triangle) extending from the bottom side of the triangle, where the x-axis intercepts the circle

Think back to basic geometry, and tell me why they wouldn't be the same angle

Remember your right triangle/perpendicular relationships @cinder sundial

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@cinder sundial Has your question been resolved?

[minor bone]

move the vectors without changing their direction

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@cinder sundial Has your question been resolved?

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@cinder sundial Has your question been resolved?

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[vapid lance]

A team consists of 17 persons including 2 captains, of which 5 are extrem patient, other 7 are extremely kind, and rest are extremely honest. A pers from the team is selected at random such that each person is equally likely to be selected, find the probability of selecting a person who is
(i) extremely kind
(ii) extremely patient or honest
(iii) extremely honest and kind​

[vapid lance]

whats the answer for the 3rd one

[vale wigeon]

!noans

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The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[vapid lance]

shudnt it be zero

cuz its and

[prime badge]

i would answer 0

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@vapid lance Has your question been resolved?

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@vapid lance Has your question been resolved?

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[rotund shoal]

How can I approach this?

Should I try to prove LHS = RHS?

[old sinew]

what is $\sin(\frac{\pi}{2}-\theta)$?

[ocean seal]

Dubs

[rotund shoal]

would it be cos theta?

so LHS = sin(theta)cos(theta)?

[naive valley]

and that hopefully brings to mind a trig identity

[rotund shoal]

ah yes

$\frac{1}{2}sin2\theta$?

[ocean seal]

Jshy

[rotund shoal]

thanks

.close

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[lilac wave]

I'd like to solve question number 3 and end up with the same result that's below it, using a method similar to the one used in the example, is that possible?

[floral palm]

In stuck in this question for a while now honestly dont even know what im doing, at first i decided simul equations out of which 2 i have shown and i need a 3rd one which i cant find. In the process if finding the third i found another way to find the area which is basically taking the area of sector for the red perimeter subtract the red triangle and the black shaded segment but i dont know any of the angles

[vale wigeon]

!occupied

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[floral palm]

These are the 2 eqs, r1 is radius for smaller circle and r2 is radius for bigger

[vale wigeon]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[floral palm]

Oh sorry