help-26

[rose horizon]

can’t handle of this

[neon iron]

for $\sin(\frac{\pi}{6})$ it's really $\frac{\sqrt{1}}{2}$

[thorny flame]

<:F_button:1095679234497843251>

[neon iron]

for $\sin(\frac{\pi}{2})$ it's really $\frac{\sqrt{4}}{2}$

[thorny flame]

<:F_button:1095679234497843251>

[rose horizon]

isn’t 1?

what

[neon iron]

30 45 60 90
Just add 1 to n for $\sin(x)$
$$\frac{\sqrt{n}}{2}$$

[thorny flame]

<:F_button:1095679234497843251>

[neon iron]

for $\cos(x)$ it's the opposite: 30 45 60 90
$$\frac{\sqrt{4-n}}{2}$$

[thorny flame]

<:F_button:1095679234497843251>

[neon iron]

yeah but the reasoning is alot more consistent when you write it as $\frac{\sqrt{4}}{2}$

[thorny flame]

<:F_button:1095679234497843251>

[rose horizon]

i see

i’m definitely failing this

thx

.close

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[proven narwhal]

Need help in finding the ratio r’/r

<@&286206848099549185>

[topaz sinew]

@proven narwhal Has your question been resolved?

[proven narwhal]

<@&286206848099549185>

[vernal vale]

try distributing everything

then, get all the terms that have R on one side

and factor it out

[flat kindle]

$\frac{r'+R}{R}=\frac{r'}{R}+\frac{R}{R}$

[thorny flame]

WhereWolf(ping if needed)

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[oak wraith]

There's 150 candies that needs to distributed amongst 5 children

1. Every child gets atleast one candy

So here I was thinking of stars and bars since both the object and boxes seems to be indistinguishable.

So my calc was just (150-5+5-1)C(5-1) = 149C4

2. AND! the oldest child gets atleast 5 candies.

Since I need to fullfill both these requirements. I was thinking of some sort of principle of inclusion exclusion?

|Oldest child gets atleast 5 candies| = 149C4 - |Where the oldest child gets 1,2,3,4 candies|

Am I thinking correctly? and how do I go further than this?

[oak wraith]

Or do I just calculate with stars and bars again? But now with 5 less candies?

[cursive thorn]

You can look at it as distributing 146 candies, yes

[oak wraith]

So just 144C4?

[cursive thorn]

No, 145C4

Each kid gets at least 1, so giving 4 in advance is enough

[oak wraith]

Your right.. He already has one 🤦‍♂️

But techincally what's wrong with this solution?
149C4 - |Where the oldest child gets 1,2,3,4 candies|

[cursive thorn]

Nothing I think, but it's more work

[oak wraith]

That's all I needed. tyvm! 🙏 ❤️

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[plush elbow]

https://discord.com/channels/268882317391429632/1160720981405487104

[drifting swift]

!noadvert @plush elbow

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@plush elbow Has your question been resolved?

[plush elbow]

this is a help channel did you even bother to look???

if you cant answer the question you dont have to respond

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[novel oriole]

Is the bearing just 15?

[pulsar sun]

Yes

[novel oriole]

Thanks

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[drifting swift]

for what purpose?

you could perhaps write it as a + (b-1)i, but we have no way of knowing if that helps at all.

!original

[topaz sinew]

Please show the original problem, exactly as it was stated to you. A picture or screenshot is best.

If the original problem is not in English, then post it anyway! The additional context might still help helpers help you. Do your best to translate.

[drifting swift]

@waxen yarrow

[topaz sinew]

@waxen yarrow Has your question been resolved?

[chilly walrus]

factor i out

a+i(b-1)

$a+ib-i = a+i(b-1)$

[thorny flame]

artemetra

[topaz sinew]

What step are you on?
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2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[drifting swift]

what the actual fuck

did you really look at my about me, see the TWO words i say not to call me, and call me by them on purpose?????

[chilly walrus]

inattentive people ig 😕

[drifting swift]

no this is not inattentiveness

this reads as just malice to me

@waxen yarrow

???

you cannot have done this on accident.

<@&268886789983436800>

[junior acorn]

relax Ann it was just a mistake

don't blow up

[drifting swift]

i don't believe it one bit

my about me literally says not to call me "bro" or "dude" and you used BOTH of those at ONCE.

[junior acorn]

just let it go, idk why you're letting some guy sitting across the planet affect you so much

[junior acorn]

how would Matrix Man know to check your bio tho

he wouldnt have randomly just read your bio for no reason

[drifting swift]

you know when you click on somebody

also

im a woman for crying out loud

[junior acorn]

why would he randomly click on your profile tho, is my question

[drifting swift]

idt it is random

i usually read the bios of those i talk to

[junior acorn]

im no one to influence your opinion, but imo "dude" and "bro" are just gender-neutral

but just let it go

[drifting swift]

"bro" or "dude" are too masculine for my liking.

[junior acorn]

i dont see why it's so bad

[drifting swift]

that's only your opinion and your experiences do not align with mine

let the record state matrix man used BOTH of these words. had he only used one, i would have been orders of magnitude less explosive and just said "please don't call me bro".

??

[fossil coyote]

don’t poke the bear

[drifting swift]

what?

did i use a big word

this feels so fucking tacky

[azure kite]

Got rid of them

[drifting swift]

thank you so much.

[azure kite]

I like to try give benefit of the doubt, the original comment I was leaning towards deliberate but still wasn't entirely sure

but that response + their weird chatgpt thank you from a while ago tipped me over

[drifting swift]

oh it was so deliberate

[drifting swift]

oh the one that was directed at me too right?

[azure kite]

Yep

[drifting swift]

ok yeah

[azure kite]

I wanted to see if they had any other interactions or comments with/about you

Gone now

:)

.close

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[fossil coyote]

well ig we close this now

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[waxen locust]

i have a question about the length between two points formula

[waxen locust]

i was thinking today that "doesn't it seem similar to if we would take abs value of the points, add together", then avoid having to square and take the square root ?

the answers are quite similar to this exercise i made for myself; 13 and 12.6

did anyone think of this before, can someone explain to me why the squaring gives better precision, yet the abs value formula is close?

wait shit

i did a mistake

i forgot to correct the y values of the normal method and now i see they are very different

and by coincidence my mistake made them seem like they yielded similar results

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[gleaming panther]

hi

[topaz sinew]

Please don't occupy multiple help channels.

[gleaming panther]

oh oops

#help-5

[urban pulsar]

close it then...

[gleaming panther]

yeah i just did

i didnt know i had one open

🙄

i know that thhis is a silly uestion but

since the 7 stays as 7 do you round it up so that it becomes 821.30?

[drifting swift]

to what precision are you rounding?

[gleaming panther]

2 DP

because i always get confused when rounding, its easy but when it says to a decimal place i get confused

cause do you keep on rounding up?

or do you just leave it as is

[craggy haven]

if you're rounding to 2 decimal places you'd leave it at two, so yes the 7 stays and you end up with 821.27

[gleaming panther]

so you dount just round it to the integer?

so 0 doesnt count as a decimal place right

[craggy haven]

nope, since it asked you to round to 2 decimal places you just look at the third to decide whether to add 1 to the second, and then you stop

[gleaming panther]

so you dont do anything with the actual 2nd one right? you dont round it with the 1st?

(and you dont do anything with the 1st)

[craggy haven]

correct

you also don't do anything with the fourth, fifth, etc

[gleaming panther]

thanks, so just the one before the question is asking you?

[craggy haven]

the one it's asking you for, it asked for 2 decimal places

821.2749
    1234

[random meadow]

remember bro if your rounding to 2dp it means there must be only two number after the decimal. so your last digit is 7 so yu look at the next number which is 4. cause 4 is below 5 it remains 7

[gleaming panther]

so only round the decimal place its asking for (like if it says round to 3 dp, you only change the 3rd dp)

[random meadow]

yes. so if it did ask to 3dp you would look at the third decimal which is 4. look at the number next to 4 and you see its 9 so the four goes up to 5, so answer would be 821.275

[gleaming panther]

Thanks

sorry for the extremely dumb question btw

.losec

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[neon iron]

im a bit stuck regarding some terminology i'd like clarification on.

1- Are "treatments" and "populations" synonymous?
2- Does "treatment combinations" mean a set of multiple populations that we are comparing?
3- Is an "experimental unit" meant to indicate the nature of one unit of the population we are considering, specifically in an experimental design model?

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[neon iron]

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[sweet shard]

There is so little math in your question

[neon iron]

this is like

the introductory section to my probability and statistics book so lol

idk where else this is meant otherwise

[sweet shard]

i'm not saying it's not something you shouldn't know. it's just something you look in a dictionary

or better in the context of your book

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[gleaming gyro]

Hello, I have a question. My teacher gave us a rational inequality to solve (x-6)(x+1)/3x < 0

My question is, is 3 still included in the interval notation?

[ionic oar]

wdym?

[steady escarp]

what was ur solution to this problem

[ionic oar]

the 3 is irrelevant

[worthy storm]

on the left side, you have a product of three things and the product must be negative; what does that tell you about how many of the things need to be negative?

[gleaming gyro]

oh

[gleaming gyro]

I didn't solve yet

how is 3 irrelevant?

[ionic oar]

Because you can multiply it on both sides

Are you familiar with solving rational inequalities with a number line?

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[sharp plover]

Combinatorics, binom coef questions
if i take 2 out of 10 then 1 out of 8, why is it a different number of combiantions than just taking 3 out of 10 (10 2) (8 1) != (10 3)

[sharp plover]

im not asking to calculate it, im asking how come this isnt true

its not mixing the orders, i figured it out

.close

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[thorn briar]

.close

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[glass horizon]

Is this correct

[wind yacht]

No

redo this

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[glass horizon]

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[neon iron]

How do I find critical poings

Points

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[remote pulsar]

how do I solve this?

[remote pulsar]

2000e^.04t

solve for t

[torpid matrix]

this isnt an equation we cant solve for anything

what does it equal?

[remote pulsar]

let me go look

im confused on how to use the equation for this kind of problem

pe^rt

[acoustic pecan]

,rccw

[acoustic pecan]

what does it mean by compounded continuously?

[remote pulsar]

its an equation used to calculate interest

[acoustic pecan]

yeah, im more looking at the 'continuously' its not like it is compounded by 4% annually?

[remote pulsar]

yeah the continuously is just and infinite amount of time it’s being compounded

[acoustic pecan]

alright, your exponent should be 1.04t then not .04t

[remote pulsar]

t would be in years, which is how many times it gets compounded im pretty sure

[acoustic pecan]

and it should be 1000 rather than 2000

you really want to be solving 1000e^(1.04t)=2000

[remote pulsar]

ohhh okay

thank you!!

.close

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[rustic sparrow]

187

[rustic sparrow]

$y=x-\frac{2+3pi}{2}$

[thorny flame]

putridplanet

[rustic sparrow]

is this right

the answer sheet says x + 2+3pi/2

[acoustic pecan]

f'(x)=-sin(x) =1
y-1=x-3pi/2 y=x+1-3pi/2

y=x+(2-3pi)/2

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[quasi stratus]

can somebody help me out with this problem

i dont get implicit differentiation

[vivid spire]

Well uh do you know what implicit differentiation is

[quasi stratus]

i dont get it at all

so

no

??

[elfin sparrow]

How did you get this answer?

[quasi stratus]

by following khan academy

-(12x^2)/(15y^2) i just simplified

[elfin sparrow]

Ok but did you do that work on your own?

[quasi stratus]

yes

but its wrong

[elfin sparrow]

By taking d/dx on both sides?

[quasi stratus]

and it says u need to find the second derivative i think thats only the first

[elfin sparrow]

Yes you're right

That's why I'm asking how you got it

[quasi stratus]

is the first derivative correct thjo

[elfin sparrow]

Because you said you don't know implicit differentiation but you probably used it to get that

[quasi stratus]

well i dont understand

it i just followed the steps

[elfin sparrow]

Yes it's the correct first derivative

[quasi stratus]

okay great

now how do i get the second one

[elfin sparrow]

Well, we can take the derivative of that using the quotient rule

[quasi stratus]

ohh

wait thats smart

im so dumb

LMAO

[elfin sparrow]

Just remember to use the chain rule when you differentiate y^2

That's what most people miss when they do implicit differentiation

[quasi stratus]

is this with the quotient rule or smth else

[elfin sparrow]

Yes, the quotient rule

But be careful when you differentiate the y^2 part

y is a function of x, so $\frac{d}{dx} y^2 = 2y \frac{dy}{dx}$

[thorny flame]

tatpoj

[elfin sparrow]

Not just 2y, but 2y(dy/dx)

[quasi stratus]

so in other words would it be 2y y`

2y y'

[elfin sparrow]

Yes

[quasi stratus]

okay

[elfin sparrow]

And you have a formula for y' right here

[quasi stratus]

in my case would it be

15y^2

so it would be uhhh

30y y'

??

[elfin sparrow]

Well you could use the simplified form with 5 instead of 15

Will save you some trouble

[quasi stratus]

oh yeah

LMAO thx

[elfin sparrow]

You also don't want to have y' in your final result, so make sure to substitute y' with -4x^2/5y^2 when you're done

Np 👍

[quasi stratus]

okay so i got

(-40* y^2 * x + 40x^2 y * y')/(25y^4)

[elfin sparrow]

Looks good, then substitute y'

[quasi stratus]

(-40* y^2 * x + 32/y)/(25y^4)

[elfin sparrow]

I think you should still have some x's with that 32

x^4

[quasi stratus]

OH

LMAO oops

[elfin sparrow]

Other than that looks good

[quasi stratus]

(-40* y^2 * x + 32x^4/y)/(25y^4)

[elfin sparrow]

Yep

It could maybe be simplified a bit but that shouldn't really matter

[quasi stratus]

its marking it as wrong

even after simplification

[elfin sparrow]

one sec

Oh

[elfin sparrow]

Both terms here should be negative

There was a negative sign in your y'

[quasi stratus]

oh

(-40* y^2 * x - 32x^4/y)/(25y^4)

[elfin sparrow]

Yeah, because y' itself had a -

didn't see that

[quasi stratus]

ohhh okay thank u sm

thats correct

[elfin sparrow]

Awesome 😎

No problem 👍

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[neat sierra]

A person looking
out the window of a stationary
train notices that raindrops are
falling vertically down at a
speed of 5.0 m/s relative to the
ground. When the train moves
at a constant velocity, the rain-
drops make an angle of 25 when they move past the window, as the drawing shows. How fast is the train moving?

[neat sierra]

i got the answer is 2.3m/s

the issue is

i dont know how to get it without being given the answer

i was playing around with 5 and tan 25 and found that tan 25 * 5 = 2.3

how would you solve this problem

I know that Vte = Vtr + Vre

By relative motion formula

Vre is given to us as 5.0m/s [down]

Vte is constant

And alpha = 25˚

But I dont know what two sides are enclosing alpha

I also don't know how to draw a diagram for this

If somebody could help me find the two above things it'd be greatly appreciated

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[pseudo wyvern]

Hi, I'm doing a uni-prep year and I can't really wrap my head around absolute values, can someone help me explain |3x+15| ≤ 0 please?

[vernal vale]

you sure you wrote it right?

[tough cargo]

Yeah there's only one solution that way

[pseudo wyvern]

yes

[vernal vale]

Okay

well absolute value is never negative

One way to think about absolute value is like ... distance

[pseudo wyvern]

so the answer is just 5 then?

[vernal vale]

it isnt

you almost got me there

[pseudo wyvern]

HAHA

okayokay

[vernal vale]

youre close

you know youre solving 3x+15=0 right

so 3x = -15 ...

[pseudo wyvern]

yeah that's what I'm telling myself

[vernal vale]

ill type it in math mode so its more clear

$3x = -15$

[thorny flame]

jan Niku

[vernal vale]

why is the answer not 5?

[pseudo wyvern]

well it's negative

[vernal vale]

yea

so, sign error

other than that, no issues

[pseudo wyvern]

alright, but where does this bring me

[vernal vale]

to -5

which should make sens

3*(-5) = -15

which 'cancels' the positive 15

leaving you 0

[pseudo wyvern]

yeah, this here is what's confusing me haha

[vernal vale]

it is designed this way

well, maybe defined is the right word

[pseudo wyvern]

but I suppose that the sum of the funny brackets just has to be 0 then cause it's not allowed to be less

[vernal vale]

no, you can do it more carefully

you dont have to assume anything

the typical, more methodical approach is to use the definition of absolute value to remove the funny brackets entirely

the result is two equations, each valid in a certain region

,w graph |3x+15|

[thorny flame]

Results provided by WolframAlpha

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

Requested by janniku

[vernal vale]

see, you can split it here

does that make sense?

I mean, the fact that you can split it

and solve it as if it were two lines instead of one inside funny brackets

[pseudo wyvern]

if i were to split these, would it be x ≤ -5 and -5 ≤ x ?

[vernal vale]

yea

conventionally, you'd leave the equals to only one line

even though theyre equal there

just to avoid overlapping ranges

if you do that in your problem, you'll only get a solution from one equation

but thats fine, we know already theres only one solution to find

[pseudo wyvern]

alright...

[vernal vale]

okay

we can apply the definition of abs value

then for x <= -5, we have |3x+15| = -3x-15

for x > -5, we have |3x+15| = 3x+15

now, you are solving two inequalities

$3x+15 \leq 0$ with $x > -5$ and $-3x-15 \leq 0$ with $x \leq -5$

[thorny flame]

jan Niku

[vernal vale]

I'm more trying to motivate that realizing the nature of absolute value is easier than doing all this I guess

although you can certainly make it more ... rigorous than just "realizing" something

[vernal vale]

in response to this

[pseudo wyvern]

alright, so in this case this abs value gives these two inequalities with the same answer?

[vernal vale]

well, it depends

the way I defined the domains for the lines, right

x > -5 and x <= -5

only one equation will give you a solution

( its the -3x-15 one, because it includes x=-5 )

[vernal vale]

your definition of the domains for the lines will give you two solutions

and theyll be the same

( they both include x=-5 )

[pseudo wyvern]

because they "split" at -5 that's the only solution?

or well, other way around

[vernal vale]

It's kind of ironic because we kind of had to solve the problem to split the domain

I think that kind of confuses the process a little in this specific problem

You could instead consider some problem like $$2 \leq |3x+15| \leq 4$$

[thorny flame]

jan Niku

[vernal vale]

then the utility of doing all this is maybe more clear

[vernal vale]

I'm not sure if that exactly answers your question

to apply the definition, you have to find the roots of the thing inside the absolute value

it just so happens thats exactly what youre looking for in the first place, in your problem

[pseudo wyvern]

yeah it doesn't haha, but I'm trying to follow best as I can.

[vernal vale]

there's another "cheat" method

[thorny flame]

jan Niku

[vernal vale]

you can write $$-b \leq f(x) \leq b$$

[thorny flame]

jan Niku

[vernal vale]

doing that here gives $0 \leq 3x+15 \leq 0$

[thorny flame]

jan Niku

[vernal vale]

maybe this makes the fact that we only have one solution more obvious ...

If you do more problems, you will see i think

the one you posted is kind of a special case

well, i think theyre not expecting you to solve it all these weird ways

so doing it is weird

sorry we got off on the tangent for a problem that isnt suited for it

[pseudo wyvern]

sorry just trying to process what I'm reading, but the reason why this abs value has only one answer is because of the 0?

[vernal vale]

Yes.

[pseudo wyvern]

if the problem was |3x+15| ≤ 4

[vernal vale]

https://www.desmos.com/calculator/wv2mh030kv

try the slider

the red region is the values of x which are a solution to the inequality

[vernal vale]

but yes, this would give you a region

[pseudo wyvern]

so a problem like |3x+15| ≤ -4 wouldn't be possible to solve?

[vernal vale]

I don't think impossible is fair

it just has no solutions.

which is fine

you see problems like $x^2+4=0$ that have no solution in algebra and stuff

[thorny flame]

jan Niku

[pseudo wyvern]

or anything really that would say that the absolute value is less than 0

[vernal vale]

yea

[pseudo wyvern]

and because of this I should be able to see from the beginning that the problem only has one solution? which would be abs value = 0

[vernal vale]

Yea

The inequality is really saying |3x+15| = 0 or |3x+15|<0

and you know right away the second one isnt gonna provide anything

if its helpful to think that way

[pseudo wyvern]

alright, thank you so much.

[vernal vale]

np

[pseudo wyvern]

this really made it much clearer, I feel like motivating the answers will be much easier with this.

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[sly cradle]

I need help with number 9 please.

[opal cliff]

derivate first

[opal cliff]

@sly cradle show ur derivate

[sly cradle]

ok

[opal cliff]

alr?

[sly cradle]

its ok i got it

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[opal cliff]

but wait, i did this very quickly bc i gtg,

[opal cliff]

it seems like its not able to derivate

at x=1

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[cyan condor]

in expansion of (1+4px)(1+qx)^n, find the values of p and q. the coefficient of x^2 is -40 and x = 0

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[humble compass]

Hello can I get help on this?

[humble compass]

Towards the bottom I have k^4/4 -2k^2

And I just used that and put it in my calculator to solve

As a fourth degree polynomial

But the answer says I should factories it so I get k^2

And my answer is different why doesn't my method work?

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@humble compass Has your question been resolved?

[winter temple]

What is the answer?

[winter temple]

2sqrt2 is equal to sqrt8

So you’re correct

[humble compass]

Oh bruh how did I miss that

Ok it's been a long day but thanks for the help lol

.close

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[charred patio]

Hello everyone. Is it true that if a shape is enlarged by an integer than it will be shifted by the distance that the scale factor is, like if the scae factor is 2 than it will be twice the distance from the centre of the enlargement, if the scale factor is a decimal than it is the opposite????

[charred patio]

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[mental surge]

hello, i am just stuck on question 1, i tried solving for the genereal solution but using estimation and stuff i still get it wrong, can anyone help me solve this?

[mental surge]

the first .close

.close

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[neon iron]

Someon echelon

[neon iron]

Help*

[shut obsidian]

Squeeze theorem on cos(x).

[neon iron]

Can u tell me what is squeeze theorem ?

Well nvm I know it

I tried it but it doesn’t work

[shut obsidian]

You know that $-1 \leq \cos(x) \leq 1$ for all $x$ in the domain of $\cos(x)$, i.e. $\mathbb R$, right?

[neon iron]

I know it already but

[neon iron]

But sin(1/x^2)?

So I will multiply cos by sin(1/x^2)?

[neon iron]

OH

[shut obsidian]

Now take the limit of the term at the very left

[neon iron]

It equal zero ?

[shut obsidian]

And of the term of the very right

Yes, both

So what does that mean for the middle?

[neon iron]

0

[shut obsidian]

Yes

You "squeeze" the middle between these two terms

Thus, "squeeze theorem"

[neon iron]

Thank you I taught I have to multiply and Dived the sin(1/x^2) by 1/x^2?

[shut obsidian]

sin(x)/x is not 1 for x -> infinity

Only for x -> 0

I think that was your idea

But that doesn't work, since we have x -> infinity

[neon iron]

Yes I see now thanks

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[neon iron]

.close

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[main terrace]

What is the tangent for f(x) = x^2 + 3x, at (1, 4)?

[main terrace]

I got 9.

[drifting swift]

you got a number?

show your work, you're definitely doing something very wrong somewhere but it's impossible to tell what just yet

[main terrace]

Ok wait

I clean and show.

Wait is it 1?

[restive inlet]

show your work

[main terrace]

I got this.

Ok.

[restive inlet]

is wrong

[main terrace]

Ok.

Wut wrong with my work?

[restive inlet]

where's your expression with the limit coming from?
what were you trying to apply?

[main terrace]

lim x -> a f(x) = f(x) - f(a)/x - a ?

Oh wait.

Oh.

I think I got it?

Lemme change and show.

Like this?

[main terrace]

Is this right?

[restive inlet]

don't forget to take the limit

also poor/lazy notation

[main terrace]

What?

I didn't get you.

[main terrace]

I didn't get this.

[topaz sinew]

@main terrace Has your question been resolved?

[main terrace]

<@&286206848099549185>

[junior acorn]

i dont get what this is doing

all you have to do is find the derivative at x = 1

[main terrace]

I still didn't learn derivates yet.

[main terrace]

This is the question.

@junior acorn

[junior acorn]

ok so

without derivatives

use the fact that the tangent only intersects the curve at one point

x^2 + 3x = mx + c

so x^2 + (3-m)x - c = 0

and because it only has one point of intersection, this quadratic has only one solution

[restive inlet]

and then evaluate this limit

also note that you're finding instantaneous rate of change using limit definitions
(essentially doing derivatives/calculus without being told that)

[junior acorn]

ohh that's what he was doing

i did not understand anything at first 😅

[restive inlet]

let me write out what you should've written

[main terrace]

Ok.

[restive inlet]

$\begin{aligned} f'(1) &= \lim_{x\to1} \frac{f(x) - f(1)}{x - 1} \
&=\lim_{x\to1} \frac{x^2+3x-4}{x - 1} \
&=\lim_{x\to1} (x+4) \
&= \text{evaluation of that limit} \end{aligned}$

[thorny flame]

ℝam()n()v

[main terrace]

So did I get it right?

[restive inlet]

depending on how strict your teacher is, you'd lose massive marks

for what you actually wrote

as mentioned, the notation was just horrid

[main terrace]

Ah.

Ok.

So the slope of tangent is 5?

[main terrace]

I am learning this on my own XD

Maybe you are the teacher...

I don't know though.

[restive inlet]

you even mentioned the use of
lim x -> a f(x) = f(x) - f(a)/x - a
albeit missing () to clearly indicate numerators and denominators,
lim x→1 f(x) → stuff
is NOT representative of that

[main terrace]

Oh.

Ok.

I should make sure that I DON'T butcher notation here.

[restive inlet]

yes

[main terrace]

Ok.

Understood.

Ok thanks!

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[neon iron]

Hi super quick one, say if I have a sequence like {6, 12, 24, 48, ...} how do I write the previous part (unsure what to call it) where it's like this...

[neon iron]

because that ^ is for an arithmetic sequence

whereas I don't know if it's the same for a geometric sequence or not?

the formula for the {6, 12, 24, 48, ...} sequence is 6(2)^n-1 where n_1 is 6.

I hope my question makes sense...

[drifting swift]

no it's just for any kind of sequence in general

[neon iron]

Ah okay, thank you!

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[tired venture]

Is this possible

[tired venture]

This was a question on my test that i redrew

Ik its physics but can anyone help me

[steady escarp]

It should be possible

You can resolve forces horizontally and vertically and get two equations

There will also be 2 unknowns; the angle of the first string and the force in the second sensor

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[tired venture]

Its a simple vector application test so im not so sure

How to solve for the 2nd sensor

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[sand pasture]

How do you prove this? f is a function from A to B and f-1(X) is defined as the set of x in A such that f(x) is in X

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[vernal barn]

q= 5 over 9(t+32)

[vernal barn]

how do i turn this

T subject of formula

[foggy bronze]

u mean

[vernal barn]

how do i make T subject of q= 5 over 9(t+32)

i forgot english for a couple of seconds there

[foggy bronze]

[cursive thorn]

$$q = \frac{5}{9}(t+32)$$?

[vernal barn]

no no

hold on

[thorny flame]

Jelle

[vernal barn]

ys

yes

[cursive thorn]

You can start by multiplying both sides by 9

[vernal barn]

then you divide both by 5 right?

[cursive thorn]

Yes

[vernal barn]

then -32

?

[cursive thorn]

Yup

[vernal barn]

alr alr thanks

so it becomes

9q over 5 - 32 = t

right?

or technically T is first

but you get the idea

[cursive thorn]

You can use '/' for division

[vernal barn]

alr

but is it correct

?

[cursive thorn]

so (9q)/5 - 32 right?

[vernal barn]

yea

[cursive thorn]

that's correct

[vernal barn]

alright thanks alot

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[neon iron]

uh

okay then

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[dapper sparrow]

Heyo, in this problem is it ok if I subtract instead of add?:

[dapper sparrow]

For ex:

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[sonic bone]

help

[acoustic pecan]

if a^b=a^c then b=c

[terse abyss]

Or, use natural logarithms

[sonic bone]

18y+15=(5y)(6y+5) because the bases equal?

and then do i distribute the 5y on the left?

[acoustic pecan]

sure

[sonic bone]

so 18y+15=30y+5?

do i distribute it to the 5 as well?

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[gusty wadi]

Really quick dumb question here, can you take a factor of x out of (x+x^(1/2))

[gusty wadi]

Like x(1+x^(-1/2))

[rose sierra]

You can. It'd be
$x(1+x^{-\frac{1}{2}})$

[thorny flame]

Ari

[gusty wadi]

Ok thank you so much

I was too tired to remember if that worked

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[neon iron]

How do I do this

Can someone help me with it idk where to start

,rotate

[stable rivet]

write out the components of the vectors as seperate things

like we know Ay = 0 right because A points in the negative x direction

and we know Ax^2+Ay^2 = 25

so Ax^2 = 25

which means Ax = -5 (cuz it said points in negative x direction)

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[neon iron]

Like this?

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[neon iron]

.clise

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[neon iron]

.reopen

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✅

[neon iron]

A Ball is shot above the horizontal at an angle of 38.9. The height of the ball is above 10m after 1.2s. What are the two velocities at these times and what are these two times?

[neon iron]

is this work for the time at 1.2s valid

wtf

i miscalculated something

my faul

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[neon iron]

A Ball is shot above the horizontal at an angle of 38.9. The height of the ball is above 10m after 1.2s. What are the two velocities at these times and what are these two times?

Is this how I’d solve for velocity at the time 1.2s?

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[acoustic pecan]

whats up

[fickle sorrel]

What is the meaning of the area being the same as the side length?

[acoustic pecan]

the meaning? also theyre not the same, they have different units

[fickle sorrel]

I find this confusing

[tawdry spoke]

One is say metre and other is sq.metre

Length is unit and area is square unit

[fickle sorrel]

They don't represent the same quantity?

[tawdry spoke]

No

Only magnitude is equal

[fickle sorrel]

1 = (∞ • 0)^2

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[grand garnet]

<@&286206848099549185>

I need help understanding how this person turned the normal volum of a cone formula im use to

To the one 2pir^3/9sqrt3

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[fickle sorrel]

Did you find the solution?

[fickle sorrel]

from the pythagoras theorem:

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[grand garnet]

I understand that but how di he find those

r^2=2/3R^2

H=1/sqrt3R

From Pythagorean theorem

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[fickle sorrel]

um

[fickle sorrel]

What is √3 and what is the meaning of h=R/√3, do you know?

This is the original problem #help-26 message

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[hardy shuttle]

Does anyone see how I can simplify the last part anymore? Or did I just do something completely wrong

[hardy shuttle]

Actually I missed one step, let me add it

Here so i'll explain my process

I first started with labeling my inside as u, and outside as y. I know to take the derivative of s(t) i need to do y' * u'. I knew what y' was but for u' I believe I needed to use product rule and quotient rule, so thats what I did on the right side. My last step was me putting it all together (y' * u')

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[uncut crow]

why is it s(t), there's no t?!

[hardy shuttle]

no idea hhaha im just treating it like a y

[uncut crow]

sry, gotta go. you might have to use some partial fractions

Lots of algebra to do

[hardy shuttle]

okay

idk how else to simplify

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[lofty moth]

Which part requires simplifying?

Is this derivatives?

[lofty moth]

Answer should be J @hardy shuttle

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[ashen rain]

If ur asked to derive that function you should simplify it first then derive it

And as what somxr said the answer should be j

Try using the fact that ln(Xy)=ln(x)+ln(y) to sumplify

Then derive normally

Also in case I didn’t know ln(x^n)=nln(x) and the cubic root is just to the power of 1/3

[frank pulsar]

https://www.youtube.com/watch?v=qDYoV_IAaJw

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[nimble cobalt]

help

[nimble cobalt]

can u help me

[west jolt]

whats your question

[nimble cobalt]

one sec

5x5 to the power of 2

[west jolt]

(5x5) to the power of 2

or 5 x 5^2?

[neon iron]

prolly the former

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[hard wolf]

this is the last question on my kinematics probelm set

I think no acceleration means they all start and end at their respective velocities

ΔA = ΔB = ΔC

Xa(Ta) - Xb(Ta) =12
Xa(Ta) - Xc(Ta) = 18
Xb(Tb) - Xc(Tb) = 8

Xb(Ta) = 6 + Xc(Ta)

ΔA = VaTa
ΔB = VbTb
ΔC = VcTc

[fluid belfry]

what does $\Delta{a}$ Xa and Ta represent

[thorny flame]

!Yajat!

[hard wolf]

Xa is position of ally

Ta is time it takes ally to finish

ΔA = ΔB = ΔC

Xa(Ta) - Xb(Ta) =12
Xa(Ta) - Xc(Ta) = 18
Xb(Tb) - Xc(Tb) = 8

Xb(Ta) = 6 + Xc(Ta)

Xa(t) = Va * t
Xb(t) = Vb * t
Xc(t) = Vc * t

VbTa = 6 + VcTa
VbTa - VcTa = 6
Ta(Vb-Vc) = 6

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[keen matrix]

I've always appreciated this explanation:

In the expression $(a+b)^{n}$, the coefficient of $a^k\cdot b^{n-k}$ is the number of ways to choose k 'a's and n-k 'b's from n pairs of $(a+b)$. For that we can choose k pairs for 'a's, and 'b's from the others. The number of ways to do it is $\binom{n}{k}$

see (https://math.stackexchange.com/a/643549/1049995)

[thorny flame]

PajamaMamaLlama

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[frank narwhal]

5x^2 + 16x + 12 = 0

[frank narwhal]

trying to factor this

5 * 12 = 60 so im pretty sure its supposed to be 6 and 10

[keen matrix]

there are a few methods, I prefer slide and divide but they're cross method or box method

[frank narwhal]

idrk them by name

[broken niche]

they have to multiply to 12 if ur saying the roots r 6 and 10

[keen matrix]

this is slide and divide

so you multiply the 12 by the 5 then get x^2+16x+60=0, can you factor that?

[frank narwhal]

then what do u do with the 5?

i dont think (x + 6)(x + 10) is right

unless it is?

[keen matrix]

that comes after your factor

now do the divide part

(x+6/5)(x+10/5)

[frank narwhal]

thats what ur supposed to do with them?

[keen matrix]

can 6 divide into 5?

[frank narwhal]

no so it stays as a fraction as 5

[keen matrix]

can 10?

[frank narwhal]

yep

2

[keen matrix]

so we get (5x+6)(x+2)

[frank narwhal]

tysm i despise factoring so much

makes so little sense to me

[keen matrix]

(we just move the 5 out in front)

[frank narwhal]

ohhh okay

[keen matrix]

you'll get better as your practice :)

[frank narwhal]

i am pretty decently sure ive gotten worse

tbh

[keen matrix]

it's an important skill as you get up into higher level mathematics

[frank narwhal]

but either way tyvm

[keen matrix]

[frank narwhal]

hopefully i'll remember this setup

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[keen matrix]

slide and divide

it's in the name so you don't forget

[frank narwhal]

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[tropic zephyr]

Help

[tropic zephyr]

I understand how to do the actual differentiation but have trouble visualizing the interval

What would be between 0 and 2pi

This what I have so far

[nimble crown]

So what do you want?

Just the numbers between 0 to 2pi?

Or when they go through the x axis

[tropic zephyr]

Based off what I know cosine is 0 at pi/2 or 3pi/2

[nimble crown]

Yea thats right

[tropic zephyr]

Now, how do I visualize that in a number line

Between 0 and 2pi

[nimble crown]

Just write it down on a number line?

Ok if it helps

Imagine the pi is gone

[tropic zephyr]

Don't know how? Would pi/2 come first

[nimble crown]

Can you plot 0, 1/2 3/2 and 2 on a number line

[tropic zephyr]

After 0

If this interval makes sense, how will I find what's between 0 and pi/2 ?

My guess it could be pi/4 or pi/6

Using the unit circle

But I'm not sure

[nimble crown]

Yea thats right

You can divide ut up more but not rlly

Sorry for late reply

[tropic zephyr]

I got from 0 to pi/2 it's positive and pi/2 to 3pi2 its negative and 3pi/2 to 2pi its negative but got it wrong

[nimble crown]

Hmmm

Now i'm unsure now, srry I dont think I am help you now

[tropic zephyr]

That's ok

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[tropic zephyr]

<@&286206848099549185>

[uncut crow]

x would be in $[0, 2\pi]$

[thorny flame]

Xenophon

[tropic zephyr]

I understand that part

Based on a later reply cosine is only 0 at pi/2 ,3pi2 or pi/2+2pin, n is a number

[uncut crow]

yep

[tropic zephyr]

I drew my interval, and something between 0 and pi/2 could be pi/4 right

[uncut crow]

to clarify this part, n is an integer, and you should have $\pi*n$ to capture $3\pi/2$

[thorny flame]

Xenophon

[uncut crow]

pi/4 is between those two numbers, yes

But how is that relevant?

[tropic zephyr]

It's concave up in that interval? I'm new to this so I'm probably wrong

Because I used second derivative

To 0

Quadrant 1 is always positive

To my understanding

[uncut crow]

The second derivative is not the same as cos(x)

You simplified to find the inflection ponts

[tropic zephyr]

I got -8cosx as second derivative

[uncut crow]

yea, so the second derivative can be negative even when cos(x) is positive

[tropic zephyr]

Oh, I was supposed to plug in to the original second derivative

[uncut crow]

Yes.

1.) Find f''

[tropic zephyr]

I plugged in to cosine itself

[uncut crow]

2.) Find inflection points: f''=0