help-13

[lapis aurora]

and then as angle <a is 42

so

<b+<c = 180-<a

which is

<b=<c=138/2=69

and as 3 equal angles it becomes

69/3

so

<ABE=<EBD=<DBC=69/3=23

is it right till here @swift terrace ?

[swift terrace]

yea

[lapis aurora]

oki

am kinda confused after that

wait

nvm

@swift terrace yo

so now i use sine law in triangle aeb

?

[swift terrace]

yea

[lapis aurora]

but

[swift terrace]

sine law

[lapis aurora]

how does it work though

i only get for 1

like

[swift terrace]

x/sin(a) = y/sin(b)

[lapis aurora]

one min

i also found out that

aeb is 115

but do u use this?

so for my triangle aeb

i got C and c

and i got A and B

C is 115*

and c is 15 units

A is 42*

and B is 23

[swift terrace]

23+42 = 65

[lapis aurora]

but dont u need 2 sides also

[swift terrace]

180-65 = 115

<AEB is 115, you are correct

[lapis aurora]

so now

if i use sin law

sine*

i only get one of them

[swift terrace]

one what?

[lapis aurora]

AB/Sin (<AEB)

which is

15/sin115

[swift terrace]

yeah

[lapis aurora]

how do i get the other one

[swift terrace]

= BE/sin(42)

[lapis aurora]

Oh yeah

OH

and we solve for BE?

[swift terrace]

yea

[lapis aurora]

cross multiplication?

[swift terrace]

nope, just multiply eeverything by sin(42)

15*sin(42)/sin(115)=BEE

[lapis aurora]

11.07

yo

no way

No way I actually did something

btw its my 1st time actually staying up till 1 just to study maths by my own heart

alr now bd

@swift terrace u take triangle abd? and do the samething?

[swift terrace]

you can

you know the angles

[lapis aurora]

alr lemme try

[swift terrace]

and have onee of the lengths

[lapis aurora]

@swift terrace

so will it be

ab/<adb =

bd/42?

wait

will it be

15/sin92 = Bd/sin42?

yes it is

lets go

@swift terrace thank you so much man

thank you so much!

Can I friend U?

[swift terrace]

sure if you like to

[lapis aurora]

I really would like 2

[cedar kiln]

@lapis aurora Has your question been resolved?

[cedar kiln]

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[oblique prawn]

Solve for x :
$4^x - 9(2^x)+8^{(log_5{7})^2} < 0$

[wraith dagger]

Programmer007

[mortal hemlock]

looks like a quadratic

[mighty shuttle]

try writinf 8 as 2^3

and then yeah, its's a quadratic

$2^x=u$

[wraith dagger]

ƒ(Why am. I here)=I don't know

[oblique prawn]

Ok but how do I solve it further:
$u^2 - 9u + 2^{3(log_5{7})^2} < 0$

[wraith dagger]

Programmer007

[mighty shuttle]

quadratic formula

[oblique prawn]

Alright, thankyou
.close

.close

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[quartz salmon]

This is a 3/4 globe. In cyl. coordinates, how can I describe the region?

I know the theta interval is π/2≤theta≤2π.

[mortal hemlock]

how is z bounded?

[quartz salmon]

It's above or equal to 0.

[mortal hemlock]

it looks like a sphere

dont you agree?

[quartz salmon]

I am not sure about the diff. between sphere and globe

[mortal hemlock]

its the same

you could also call it a ball

or a 3d circle

[quartz salmon]

I see

So are we meant to use a formula here?

[mortal hemlock]

no?

do you know the equation for a sphere?

[quartz salmon]

Not really

[mortal hemlock]

do you know it for a circle?

[quartz salmon]

x^2 + y^2 = r^2

So we just add a third dimension, z?

[gloomy crag]

/help@wraith dagger

[quartz salmon]

@mortal hemlock

[mortal hemlock]

yep

for a sphere its x^2+y^2+z^2=r^2

[quartz salmon]

Alr

[mortal hemlock]

so we have a condition for z

[quartz salmon]

So we have x^2 + y^2 + z^2 = 4?

Or is it radius^2?

[mortal hemlock]

is the radius 4?

or is the radius2?

its radius^2

[quartz salmon]

radius^2 means it'll be 16

[mortal hemlock]

we can rewrite it to $z^2=r^2-x^2-y^2 \implies z=\sqrt{r^2-x^2-y^2}$

[wraith dagger]

Flappie

[mortal hemlock]

for the top half of the sphere

[quartz salmon]

The answer says it's 4^2 for the radius

And there's no y part…

[mortal hemlock]

does it not depend on x and y aswell?

isnt it a cut off sphere?

[quartz salmon]

I would assume so

It's a 3/4th sphere

[mortal hemlock]

yeah, so it should also be dependent on x and y

unless thats baked into r

like r(z) type beat

[quartz salmon]

Yes

It'd make sense if it was so

So we get x^2 + y^2 + z^2 ≤ 16

But how do they arrive at that inequality?

[mortal hemlock]

ohhh, they do make r dependent on x and y

[quartz salmon]

I can agree with z≥0

[mortal hemlock]

r^2=x^2+y^2

if you plug that in

you get the same thing

[glad peak]

They use a change of coordinates to cylindrical coordinates

With
x = rcos t
y = rsin t
z = z

Then plug into sphere eq

[quartz salmon]

Yes but how do we come to the conclusion that it's less than or equal to 16?

Is it because all points (x, y, z) of the sphere are less than 16 since thats the radius?

[glad peak]

Yes

[quartz salmon]

That was a bad explanation but ys you got the point, good

[quartz salmon]

How do we go from expressing it in those terms to [r^2+z^2 \leq 16]

[wraith dagger]

dghf

[glad peak]

Plug these in

[quartz salmon]

Since I am not seeing sin or cos

[glad peak]

What do you get after subbing

[mortal hemlock]

they drop out to become 1

[quartz salmon]

I get rcos^2 + rsin^2 + z^2 ≤ 16

r (cos^2 + sin^2) + z^2 ≤ 16

[mortal hemlock]

almost there

[glad peak]

the r is also squared

[glad peak]

But this is the correct next step

[quartz salmon]

i see now, alright thanks

.close

[cedar kiln]

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[marble sluice]

I was checking my answer for an integration problem and it’s hard for me to get the derivative for this $$ -\frac{xcos 10x}{10} \frac{sin 10x}{100} + c $$ what should I do? I know the product and quotient rule but keeping track of everything is overwhelming.

[wraith dagger]

Nerdy_Coder

[crimson delta]

you dont need any quotient rule here

[tranquil tartan]

f:g:h g is 60 percent more than h , f is a third of g , simplify f:g:h

[dim gulch]

go open a new help channel in available math help

its one category above this one

[marble sluice]

How?

[crimson delta]

10 and 100 are just numbers

[marble sluice]

So what?

[crimson delta]

you can just move them out of the derivative

same thing as how the derivative of 2x^2 is the same as 2 times the derivative of x^2

[marble sluice]

oh

Dena should I go back and study derivative problems?

I keep struggling hard with integration because I suck at keeping track of things

[crimson delta]

then yes that would probably be a good idea

doing some more involved chain rule/product rule/... things is a good way to train getting better at keeping track of stuff

[marble sluice]

thanks

[cedar kiln]

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[frank phoenix]

can someone help me out with this one? CALCULATE the value of m in the equation 2x^2 – 5x + (3m – 2) = 0 so that the product of the roots
be 2/5

[austere ember]

vieta's formulas tell you that for a quadratic ax^2 + bx + c, the product of the roots is given by c/a

so in your example, figure out what a and c are, plug those into c/a to get the product of the roots, set that equal to 2/5, and solve that equation for m

[cedar kiln]

@frank phoenix Has your question been resolved?

[frank phoenix]

thanks bro

[cedar kiln]

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[robust sage]

Can someone please give me a clue on where to begin with this

[cedar kiln]

@robust sage Has your question been resolved?

[hallow junco]

I would try something like induction, but I am not sure

(I am so unsure that I am actually surprised that it speaks about the n+2 derivative of y despite we only supposed y n-times differentiable)

[robust sage]

I was thinking about an induction route, but I’m not sure how Leibniz rule comes into play

[hallow junco]

Try maybe isolating y in the hypothesis and then differentiating the other side n times

[cedar kiln]

@robust sage Has your question been resolved?

[cedar kiln]

@robust sage Has your question been resolved?

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[long mauve]

how do I do this

[cedar kiln]

Please don't occupy multiple help channels.

[flint plinth]

maybe try factoring the first one as x(x^2 + y) = 0, then replace x^2 with y^2

see if that takes you anywhere

[silent finch]

or take cases on whether x=y or x=-y

[azure swift]

In general cases is the way to go, you could even from the eq. Bungo suggested, separate as cases for when x=0 or y=-x^2

[cedar kiln]

@long mauve Has your question been resolved?

[long mauve]

that was my first idea but I have no clue what to do with that

I just keep getting that it's 0 and that's it

[cedar kiln]

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[long mauve]

how do I prove f(x) = x^2 + 4x - 5, f: [-1,2) -> [0,7) is injective

[long mauve]

I tried using algebra to get f(x1) = f(x2) => x1 = x2 but I got stuck at x1^2 + 4x1 = x2^2 + 4x2 and didn't know what to do after that

[azure swift]

Try and complete the square

And then observe that you’ll have unique roots due to your domain

[long mauve]

what does completing the square mean

[azure swift]

so for example x^2 + 4x = 0 can be rewritten as

x^2 + 4x + 4 = 4

Hence “completing the square”

Since the LHS can now be naturally factored

[long mauve]

I don't really understand

[azure swift]

Does x^2 + 4x + 4 not look familiar?

[azure swift]

Indeed

[long mauve]

how does that help me I don't really get

(x1 + 2)(x1 + 2) = (x2 + 2)(x2 + 2)

I can get rid of x1+2 and x2+2?

to get x1 + 2 = x2 + 2?

[azure swift]

Yep

[long mauve]

ah I see, then -2 and I get x1 = x2

[azure swift]

Since (x+2)^2 will trivially be injective on that domain, or viewed differently the square root will give a unique root

Yes!

[long mauve]

what does that mean

I see it as just a way to rewrite the same equation

[azure swift]

If you had a different domain, say the whole real line; you wouldn’t be able to draw the same conclusion

Basically the equation x^2 = y is not unique for all x on the real line

[austere ember]

you can also show this in a simpler way

you can prove that x1 != x2 -> f(x1) != f(x2)

[azure swift]

But that uses the exact same reasoning

[austere ember]

actually yeah just go with this approach

[long mauve]

with that I'd still be confused on what to do with the 4x

[austere ember]

well

f(x1) != f(x2) means that (x1)^2 + 4(x1) != (x2)^2 + 4(x2)

i'm going ahead and cancelling the 5

that's the same as trying to show that (x1)^2 - (x2)^2 + 4(x1 - x2) != 0

clearly 4(x1 - x2) is nonzero since x1 != x2

i think there's a way to show that the whole thing will be nonzero as well

[long mauve]

I'll stick with the first way, I feel more comfortable with it

thanks guys

[cedar kiln]

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[elfin prism]

A lift of mass 200kg is moved up and down by connecting a motor, the maximum tension beared by a wire is 7000N a person of mass 50kg wishes to find the maximum acceleration, standing on a balance placed in the lift he observes its reading when it is moving up. he observed a maximum reading of 65kg, find the maximum acceleration

[vague rapids]

hari putar

harry potter

[vague rapids]

🧙‍♀️

harry pottress

[elfin prism]

can u help w the sum

[vague rapids]

idk

[elfin prism]

https://tenor.com/view/poop-cat-meme-dark-fnaf-gif-2491098315710587638

[elfin prism]

<@&286206848099549185>

[merry rivet]

this is applied physics lol, im a pure math student

wont be able to help

[elfin prism]

🥲 its alright sorry for the inconvenience

[vague rapids]

alright

so

f=ma

but there is an upward accelaration as well

[elfin prism]

yeahh

[vague rapids]

so you have f=ma-mg

since a and g have opposite directions

[elfin prism]

if the lift is accelerating upwards the apparent weight is larger right soo it should be m(g+a) right

[austere pawn]

yea

I think so

[vague rapids]

oh alright

That's a pseudo force

considering the observer inside the lift

the apparent accelaration is downwards

so you have T=ma+mg

where m is the combined mass of lift and the person

original weight or person: mg=500
new weight: m(g+a)=650

[elfin prism]

yess

what happens to the tension of the lift

[vague rapids]

that should give you the maximum acceleration

[elfin prism]

3m/s^2 ?

[vague rapids]

maximum tension is 7000N
so, T_max=(M+m)(g+a)

(M+m)(g+a)=(250)(13)<7000

so it seems that the given information about maximum tension is unnecessary

[elfin prism]

ohh

jeez

and so if we were to find the maximum number of people that could travel in the lift without breaking the wire, how could we do the calculation

the mean mass is 50kg

[vague rapids]

this formula right here

(M+xm)(g+a)=T_max=7000

x is number of people and m is mass of the person

[elfin prism]

M is mass of the lift?

[vague rapids]

yes

[elfin prism]

Thanks a lottt ❤️

[vague rapids]

also are you Indian

[elfin prism]

Sri lankan

[vague rapids]

This emoji lead me to that statement

[elfin prism]

.close

[cedar kiln]

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[glass sky]

How can I show that, if $g^{(p-1)/q} \not\equiv \pmod{p}$ for every prime divisor $q$ of $p-1$ then $g$ is a primitive root mod $p$?

[wraith dagger]

Edlingem

[hallow junco]

Suppose that g is not a primitive root and see what are the consequences (using Fermat's little theorem/Lagrange's theorem)

[glass sky]

What is Lagranges theorem?😅

[hallow junco]

Oh, you did not follow a group theory class ? (Sorry, there is no need to know Lagrange's theorem)

It is a generalization of Fermat's little theorem

[glass sky]

No, this is an exercise from elementary number theory

[hallow junco]

Okey, sorry. Do you know Fermat's little theorem ?

[glass sky]

Yes.. a^p = a mod p right?

[hallow junco]

Yes, that's it

[glass sky]

okey

[hallow junco]

But there is a precision

You probably know about invertible elements mod p, if the exercise is about primitive roots ?

[glass sky]

yes

[hallow junco]

Perfectly, so you have the tools to solve the exercise :

• g is (supposed ?) invertible, so use it to deduce a more suitable equation for the exercise starting from Fermat's little theorem
• Suppose g is not a primitive root mod p, and see what happens
• Conclude

(If it is not enough, you may ask for more details)

[glass sky]

Yes invertibility is assumed as p and g are coprime 😄

forgot to mention it

[hallow junco]

Perfect, I had little doubt but we never know xD

[glass sky]

But little fermst requires prime in the exponent but I only got this p-1/q and p-1 as modul

[hallow junco]

Yes, this is true, but g is invertible, so from g^p = g (mod p), you may deduce something similar to (a "special case" [for q not prime] of) g^(p-1)/q = 1 (mod p)

[glass sky]

by multiplying by the inverse i'd get g^p-1 = 1 mod p

So thats basically Euler-Fermat isnt it?

[hallow junco]

Yes, that's it. From there, suppose g is not a primitive root mod p

[hallow junco]

Possible, I have difficulties remembering every name 😅

[glass sky]

So there must exist a l which divides p-1 and satisfies g^l = 1 mod p where l is the order of g

g would be primitive root if no such l < p-1 exists

[hallow junco]

Exactly

And you can use the fact that l divides p-1 to conclude

[glass sky]

you mean by „cancelling“ the l out in the exponent?

[hallow junco]

No, look at the prime decomposition of l

[glass sky]

this would be a subset of the prime decomposition of p-1 isnt it

[hallow junco]

Exactly

Took one of the prime in this list, and check what happens

[glass sky]

excuse me?😅

[hallow junco]

Let q be a prime divising l

[glass sky]

then q also divides p-1..?

[hallow junco]

Yes

But that is not the point (or it is, sorry xD)

What is (p - 1)/q expressed in terms of l ?

(Write for example l = qm)

[glass sky]

some multiple of n divided by q?

[hallow junco]

Not sure to understand

[glass sky]

p-1 is some multiple of l

because l divides p-1

[hallow junco]

Not sure to understand xD, but what I meant with that was writing (p - 1)/q = (p - 1)/l * l/q

Does it help to conclude ?

[glass sky]

this would contradict because p-1/l * l/q is less than l which was said to be the order right?

So it cannot be congruent to 1

[hallow junco]

No, it is greater in general

Wait

What

(Sorry xD)

I am not sure to follow everything, but no need to use contradiction

[glass sky]

okey

[hallow junco]

Ooooh, I am sorry

I did not read carefully enough

[glass sky]

?

[hallow junco]

Here, we are looking for expressions of the type g^(p-1)/q. So, instead of considering the prime decomposition of l, consider instead the decomposition of (p - 1)/l

(And this is not non sense, because you have g^(p-1)/((p - 1)/l) = g^l = 1 )

I am sincerely sorry for my mistake

[glass sky]

no problems😅 thank you for helping 🙂

[glass sky]

i cannot follow.. what does the prime decomp. tell me?

[hallow junco]

"With pleasure" (Do we say that in english ?)

[hallow junco]

(p - 1)/l

No wait

What am I doing

[glass sky]

don't know tbh i am from austria 😅

[hallow junco]

Forgot my last messages xD

[glass sky]

so the initial approach was right?😅

[hallow junco]

So, what is my idea :

I will do it explicitly so I do not make mistake again

I am sorry, I hate do that

Since you have l < p-1 such that g^l = 1, and since we know that l divides p-1, you have a number m > 1 such that g^(p-1)/m = 1

With m = (p - 1)/l

Why am I doing this ? In order to have an expression similar to g^(p-1)/q = 1

Up to here, is it ok ?

[glass sky]

yes 👍

?

[hallow junco]

Not really pedagogical and I need to go. I removed to let the place to others if you need more help

[glass sky]

oh okey

[hallow junco]

Sorry

[glass sky]

thanks for your help

np

[hallow junco]

(I sent in pm a hidden screen of my last messages, in case nobody take over)

[glass sky]

Thanks 🙂

[cedar kiln]

@glass sky Has your question been resolved?

[cedar kiln]

@glass sky Has your question been resolved?

[cedar kiln]

@glass sky Has your question been resolved?

[cedar kiln]

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[mossy nova]

whats wrong with this

[mossy nova]

that purple function cant go higher 0.5

[vagrant elbow]

What are you trying to do?

[cedar kiln]

@mossy nova Has your question been resolved?

[mossy nova]

interpolate between 0 and 1

[vagrant elbow]

Well

I don't think I can help you

But are you sure a is meant to be 0

[mossy nova]

a can be 0

[floral arrow]

I(x) already does that, what is xI(a) supposed to do?

[mossy nova]

i was dumb

i figured out my error

.close

[cedar kiln]

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[boreal linden]

If the range of y= x^2-x-2/x^2-5x+6 is R- {a,b}, then Mod a + mod b =?

[boreal linden]

kindly help me with the above sum

<@&286206848099549185>

[steel heart]

what do you think should be the two real numbers who shouldn't be there

a general trick involves that denominator≠0

[boreal linden]

wdym

[floral salmon]

$y = \frac{x^2-x-2}{x^2-5x+6} = \frac{(x-2)(x+1)}{(x-2)(x-3)}$\
Look for removable discontinuities, asymptotes and anything else that would not allow you to achieve certain $y$ values. As a hint there are 2 of these $y$ values, namely $a$ and $b$.

[wraith dagger]

Crystopher

[boreal linden]

i did till this step and idk how to proceed

i've cancelled x-2 in numerator and denominator

[floral salmon]

be careful when 'cancelling out' terms, you are making a certain assumption by doing so, namely that $x\neq 2$, else youu would be dividing by 0. But for investigating asymptotes it may still be a good idea to simplify as you did, why would this be the case? Do you know how to find asymptotes?

[wraith dagger]

Crystopher

[boreal linden]

no idk-

[floral salmon]

what do you know about the topic of asymptotes?

[boreal linden]

nothing tbh

[floral salmon]

you should begin by getting familiar with what these are, as you need to know what they are and how to find them (specially horizontal asymptotes), consult material you may have at your disposal. If you want further help you may wait for someone else to help you since I will be leaving shortly. 15 minutes after the last message in this channel you are allowed to ping helpers to ping helpers out there. Good luck.

[cedar kiln]

@boreal linden Has your question been resolved?

[boreal linden]

<@&286206848099549185>

[wraith lagoon]

hi

[cedar kiln]

@boreal linden Has your question been resolved?

[boreal linden]

If the range of y= x^2-x-2/x^2-5x+6 is R- {a,b}, then Mod a + mod b =?
can you explain it by anychance?

[boreal linden]

nvm

.close

[cedar kiln]

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[cedar kiln]

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[bitter flint]

Hello

[bitter flint]

Can someone make this question an actual coded equating

[steel heart]

what is on the RHS?

!original

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[crimson sedge]

what

[bitter flint]

Can u make this question like generated into the computer

like I want the question to be more clear

@crimson sedge

[steel heart]

well i want the question to be more clear too

[mortal hemlock]

you mean like this? $-\frac52+\frac56=0$?

[wraith dagger]

Flappie

[bitter flint]

Yes thank you

dam atleast u understand

Like idk how to explain but that’s the best way

[steel heart]

outrageously false claim

[bitter flint]

.close

[cedar kiln]

Channel closed

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[steel heart]

that's not true lamo

*lmao

nvm

[mortal hemlock]

go to #latex-help and do it yourself next time

your question was super confusing and made no sense

and it was in the wrong channel

[bitter flint]

Oh shit

I just realized that

[bitter flint]

how do I do it myself

[steel heart]

$\text{insert LaTeX here}$

[wraith dagger]

Pro_Hecker

[steel heart]

lol

two dollar signs

$x^223424 - x^3483$

[wraith dagger]

Pro_Hecker

[cedar kiln]

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[bitter flint]

I need help on this

[steel heart]

this example might help

[bitter flint]

I understand the cross multiply stuff but the LCM stuff

It’s just weird

and idk what I’m suppose to find the LCM of

[steel heart]

cross multiplication and LCM will give the same denominator if denominators we're adding are co-primes

[bitter flint]

Do I have to find the LCM or do I just multiply

?

@steel heart

[steel heart]

both will give the same answer

[bitter flint]

So i don’t have to use LCM

@steel heart

[steel heart]

In this case, doesn't mattere

[bitter flint]

in this case?

[steel heart]

yes

[bitter flint]

So what do I find the LCM of?

@steel heart

[steel heart]

denominators

[bitter flint]

Both of them only

And cross multiply

[steel heart]

yes

[bitter flint]

that’s it

[steel heart]

no

[bitter flint]

Alright

[steel heart]

only cross multiplication or LCM not both of 'em

[bitter flint]

Oh

So in this I only use LCM

[steel heart]

ok

[bitter flint]

And leave the numerator

[steel heart]

no, you need to multiply it with some number

[bitter flint]

With what?

[crimson sedge]

cross multiply is pretty much lcm

[bitter flint]

I’m so confused

what do I do in this case?

@crimson sedge

[steel heart]

sorry for confusion, I am not a good elementary teacher

[crimson sedge]

bro im here u dont have to ping me

[bitter flint]

Ok

[crimson sedge]

whats the smallest number that both 5 and 3 go into

as multiples

[bitter flint]

1

[crimson sedge]

no

1 goes into them

were looking for numbers 3 and 5 go into

[bitter flint]

5 is odd

No numbers?

0

[crimson sedge]

oml

[bitter flint]

???

6 and 10

Then

[crimson sedge]

when does 5 go into 6????

?

[bitter flint]

Alright then just 10

[crimson sedge]

when does 3 go into 10?

[bitter flint]

15

There

[crimson sedge]

yes

now make 15 the denominator of both the fractions

[cedar kiln]

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[cedar kiln]

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[soft lark]

how am i supposed to know what will be on a intersection

before i calc it

[cedar kiln]

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[urban salmon]

Yo all the math pros here, help me with this 3 please, really appreciate it!

[dull oxide]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[urban salmon]

2

[dull oxide]

!show

[cedar kiln]

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

[urban salmon]

i have got the answer for the first one but stuck on the last 2

[dull oxide]

For #2, find angle ACD or CAD

[urban salmon]

find the length of x

[dull oxide]

Yes. I'm telling you how to find x

[urban salmon]

go ahead

[dull oxide]

☝️

[urban salmon]

oh alr

like find the triangle ACD? or just one angle of it

[dull oxide]

either works

when you find either angle, you can sohcahtoa to solve for x

[urban salmon]

alr thx

[cedar kiln]

@urban salmon Has your question been resolved?

[austere pawn]

if you are done, .close

[cedar kiln]

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[fallen oxide]

Can someone help me set up the bounds for this triple integral / confirm that I am doing it right? I've got 0<=z<=y, 0<=y<=1, -1<=x<=1

[fallen oxide]

Mainly working from analysis of this graph

But I am worried that it is wrong, because I never directly use the y = 1 - x^2

.close

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[elder pine]

i dont know how to do the 3rd one

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[dull oxide]

$7.\overline{2}=7.2222222...$

[elder pine]

1

[wraith dagger]

SWR

[elder pine]

how do i apply this to solve the 3rd one tho

i think so ?

its 10.022

yeah but why do i have to drag the 7.2 out

to 7.222

why 2 more times tho

ok i get it

thanks

.close

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[elder pine]

how do i get 0.8333

[austere ember]

can you write 0.3 repeating as a fraction?

and can you write 2.5 as a fraction?

[elder pine]

0.3/1

2.5/1

[austere ember]

it's 0.3 repeating, the 3 goes on forever

[austere ember]

that kind of feels like cheating lol you can take any number and call it a fraction by dividing by 1, i mean maybe can you write it as a fraction where the numerator and denominator are whole numbers?

for example if i gave you 0.5, i could say 0.5/1 but that's also the same thing as 1/2

[mortal hemlock]

$0.\overline{3}=0.3333....\neq 0.3$

[wraith dagger]

Flappie

[austere ember]

^^^ 0.3 repeating is not the same number as 0.3, the 3s after the decimal point go on forever in the first one

[elder pine]

yeah but what exactly do i put in my calculator to get 0.8333?

i got 0.825

[austere ember]

well if you have a calculator you can't really enter 0.3 repeating since there's an infinite number of 3s

you can either spam a bunch of 3s to get an answer that's close enough

or you can do this by multiplying fractions

which you can actually do by hand

[elder pine]

ok i got it thanks

.close

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[uncut tulip]

could someone please help me.. this is so confusing for me

[uncut tulip]

well i had got 50x^9 - 105x^6

but i dont see that on the answer chioces

[cerulean star]

you can distribute the LHS factor and take the derivative of each term, like usual, and that will be the same answer you should get with Product Rule.

[uncut tulip]

the problem switched up on me

[cerulean star]

it's exactly the same pattern though

hmm these answers do look strange

[uncut tulip]

i get 72x^7-162x^5

but like

where do i put it

[cerulean star]

I got
72x^7 - 162x^5

[uncut tulip]

ah

[cerulean star]

so...

[uncut tulip]

but like where would it go

yk

[cerulean star]

what are they doing here

oh

they want you to use the product rule formula

and not cheese it

by relying on Power Rule alone

[uncut tulip]

wdym

[cerulean star]

^

that formula

let u = 9x^5 and let v = x^3 - 3x

(both u and v are functions of x)

then

(use the formula)

they don't want you to simplify it in part a

try it

[uncut tulip]

hm

i dont get it

could u show me please

[cerulean star]

do you know what dv/dx means?

@uncut tulip

[uncut tulip]

nope

[cerulean star]

it's Leibniz notation for the derivative

dv/dx := "The derivative of v with respect to x"

(it's just the derivative)

are you able to compute that?

[uncut tulip]

not really im really confused rn

[cerulean star]

you were able to calculate the derivative of each term a few minutes ago...

[cerulean star]

^

type the derivatives of u and of v

dv/dx = v'
du/dx = u'
Leibniz notation is more useful than "prime notation", you will understand that soon

but they mean the same thing here

@uncut tulip

[uncut tulip]

man this is reallly confusing me out now

[cerulean star]

you
just
did
this
yourself
a
few
minutes
ago

it's the same thing as before

it's just taking the derivatives of polynomial terms

"power rule"

[cerulean star]

^

[gritty viper]

They're forcing you to use product rule

[cerulean star]

you found the derivative of 9x^8 - 27x^6

[cerulean star]

check the chat man, we're past that

[cerulean star]

^

@gritty viper you can have this one if you want

[gritty viper]

Do you have any lecture notes on product rule / how did you learn it

[uncut tulip]

we dont get lecture notes and the ones i have are at home and im not at home 😦

[gritty viper]

Do you know what u' and v' are

[uncut tulip]

can u show me it done plz

[gritty viper]

no

!nosols

[cedar kiln]

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.

@uncut tulip Has your question been resolved?

[cedar kiln]

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[drowsy rapids]

anotha simple question here

[drowsy rapids]

does it matter if i did A(x+2) and B(x-2) instead of whats shown

still with their denomiators ofc

[drowsy rapids]

<@&286206848099549185>

[humble karma]

It's likely going to swap the values you'd get for A and B, and in general the result won't be equivalent to the original fraction you had to separate. The important bit is that you set one variable for x+2 and one for x-2 and stick with it.

[mortal hemlock]

A and B are just variables, you can rename them to T and P and nothing would change

[drowsy rapids]

.close

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[drowsy rapids]

.reopen

[cedar kiln]

✅

[drowsy rapids]

im back

so where does the A = 3 and B = 2 come from

i see how the A + B = 5 and -2A + 2B = -2

[humble karma]

If you start with the first equation, you can rewrite that as A = 5-B right ?

[drowsy rapids]

yea

[humble karma]

What if you plug this inside the second equation?

Could you solve for B then?

[drowsy rapids]

yes you could....

i see

https://tenor.com/view/spongebob-patrick-brain-dead-drool-bored-gif-7233020

me w these questions

thanks lol

[humble karma]

No worries

[drowsy rapids]

.close

[cedar kiln]

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[drowsy rapids]

.reopen

[cedar kiln]

✅

[drowsy rapids]

how does it want it put?

[cedar kiln]

@drowsy rapids Has your question been resolved?

[drowsy rapids]

<@&286206848099549185>

[humble karma]

Hmm. Maybe they just want $\frac{A}{x+2} + \frac{B}{x+9}$?

[wraith dagger]

Azyrashacorki

[drowsy rapids]

https://tenor.com/view/rebelnred-so-smart-gif-21417606

.close

[cedar kiln]

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[wet inlet]

Don't know how to complete 2 b here, can't tell if its 7/15 x 7/15 or something else

[cedar kiln]

Please don't occupy multiple help channels.

[wet inlet]

Why isn't my other one closed lol wtf

[floral arrow]

No replacement means the number of marbles in the bag is different the second time

[wet inlet]

Oh 7/15 x 6/14 ?

[floral arrow]

Yes

[wet inlet]

Thanku

.close

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[wet inlet]

.reopen

[cedar kiln]

✅

[wet inlet]

Nevermind I confused myself

I don't understand what the final answer is

Its not 42/210 is it

[floral arrow]

It is

Though you can simplify

[wet inlet]

Okay sweet I gotta simplify it

1/5 yes?

210/42 > 5

[floral arrow]

Yes

[wet inlet]

Thankuthanku

.close

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[sand cradle]

"There are no primes that can be written as the difference of non-consecutive squares of natural numbers".

[sand cradle]

What's the premise and what's the claim here?

Premise: n is a prime. Claim: n can't be written as the difference of ...

Or

Premise: n is the difference of ... Claim: n is not a prime

Why not the latter?

.close

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[limber geyser]

i dont get why we are using the present value eqn in this example

isnt the question basically asking what the total of everything will be after 12 months

so shouldnt i use

this eqn which finds the total amount

instead of this one which is just the present value

[cedar kiln]

@limber geyser Has your question been resolved?

[cedar kiln]

@limber geyser Has your question been resolved?

[cedar kiln]

@limber geyser Has your question been resolved?

[limber geyser]

<@&286206848099549185>

[low dragon]

hello mate

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[limber geyser]

thx for responding, the explanation is above

[low dragon]

ok imma be honest im exhausted right now

but imma get someone else to help u

cos i can see you have been waiting a while

@alpine lark

[limber geyser]

thx sm

[cedar kiln]

@limber geyser Has your question been resolved?

[warped coyote]

So this looks like an intuition question more than a math question; why do we value the machine in terms of the initial payment minus discounted payments instead of discounted payments minus initial payment. The simple answer is: almost all of the time you want to use pv not the annuities formula

To simplify this problem consider n=1 t=1 r=7% R=1

Or just consider this

What does the annuities equation tell us?

The fair value of a contract that pays that amount per year (compounded), not the value of an asset that is discounted at a given rate

@limber geyser

[limber geyser]

okay thans sm for the help

sorry for responding so late

.close

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[snow summit]

did i get this one right?

[ancient lodge]

yeah

[void sand]

[snow summit]

ty

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[snow summit]

.close

[cedar kiln]

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[cedar kiln]

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[gusty charm]

im having trouble with this question

[void glen]

let’s go step by step

given that the parabola is opened downwards

would a be 1 or -1?

[gusty charm]

-1

[void glen]

yep

do you know what the form of the parabola they gave you there is called?

and more so

what h and k represent

[gusty charm]

Write an equation for the graph in the form y=​a(x-h)squared + k, where a is either 1 or minus 1 and h and k are integers.

thats all they gave me

[mortal hemlock]

https://www.desmos.com/calculator/htov42fsss try playing with this graph a bit, it might help with your understanding

[void glen]

the form they gave you there is called vertex form

where h and k are the coordinates of the vertex

h is the x coordinate and k is the y coordinate

and yeah if you still don’t understand, i would recommend playing around in desmos like flappie suggested

[mortal hemlock]

palying around with these kinds of graphing calculators is a great way of getting a good intuitive sense of what a function does

[cedar kiln]

@gusty charm Has your question been resolved?

[gusty charm]

i used the graph and i got y=a(x-h)squared + k but it was wrong

[gusty charm]

<@&286206848099549185>

!.close

.close

[cedar kiln]

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[cedar kiln]

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[true wharf]

For question D (in blue)23 have to simplify. Are both answers correct if my only task is simplifying

[runic garnet]

,calc 7^4

[wraith dagger]

Result:

2401

[runic garnet]

@cedar kiln both answers are correct?

[wraith dagger]

Result:

true

[runic garnet]

FALSE

bot is wrong

@true wharf

the first answer is incorrect

[true wharf]

Why is the first answer incorrect I used the law of exponents rule

[runic garnet]

$(ab)^c = a^c b^c$

[true wharf]

?

I just used the negative exponent law

[runic garnet]

$(-7w)^4 = (-1)^4 \cdot (7)^4 \cdot (w)^4$

[runic garnet]

the exponent distributes to the -7, as u did in the second answer

get it @true wharf ?

[true wharf]

One sec lemme see

So I have to distribute the exponent before I am even able to do anything else?

So basically the negative exponent law doesn’t apply here?

[runic garnet]

the negative exponent law is this?

[runic garnet]

that?

[true wharf]

This one

[runic garnet]

ok

yea it applies

[true wharf]

What would the “a” value be?

Would it be -7w or just w

[runic garnet]

-7w

cuz the -7 is inside the parentheses under the ^4 exponent too

its not just the w

[true wharf]

Okay so why is

-7w^-4 wrong?

Just based on the exponent law

[runic garnet]

because it should be (-7)^-4 too

$\frac 1{(-7w)^4} = (-7w)^{-4} = (-7)^{-4} \cdot w^{-4}$

[true wharf]

I see

What is -7^-4?

By itself

[runic garnet]

1/2401

[true wharf]

OHHH

so they’re equivalent

[runic garnet]

its (-7)^(-4) btw

[true wharf]

I just didn’t finish simplifying it right

[runic garnet]

not -7^-4

[true wharf]

Oh gotcha

[runic garnet]

no

-7 needs the -4 exponent too

in ur first answer, -7 didnt get the -4 exponent

only w did

[true wharf]

I see because you have to distribute the -4

Exponent

[runic garnet]

yes exactly

[true wharf]

Ok wait

So this is what I have so far; I understand to this point

However I want to know what the next steps are to simplifying it

I would multiply (-7)^-4 correct?

Then (w)^-4

[runic garnet]

multiply? eh

well

u could do whatever order u want

but yea

[true wharf]

So this works?

[runic garnet]

ye

[true wharf]

The 4 is an exponent btw in the denominator

Ok got it thank you

.closed

[runic garnet]

without the d

[cedar kiln]

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[sand patio]

I feel like i did something wrong when integrating but I cant seem to figure out what?

[cedar kiln]

@sand patio Has your question been resolved?

[sand patio]

.close

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[lilac tusk]

hi i want a help

[royal matrix]

Just ask your question

[mighty shuttle]

[cedar kiln]

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[uncut mason]

first thing to consider is dividing both sides by t^2 to make the integration factor clear. dQ/dt + 1/t^2 Q = 2tQ and then you must subtract 2tQ from both sides. dQ/dt + (1/t^2 - 2t) Q = 0. it's now clear that the integration factor is e^(∫(1/t^2 - 2t)dt).
Next step is to compute ∫(1/t^2 - 2t)dt. As 1/t^2 = t^-2, we have that
∫(t^-2 - 2t)dt = ∫(1/t^2 - 2t)dt
= -t^-1- t^2.
So we now have that e^(-t^-1- t^2) is the integration factor.
d/dt (e^-(t^-1- t^2)Q) = 0 <- this statement derives from the fact that when the product rule is applied to (e^(-t^-1- t^2)Q) we get e^(-t^-1- t^2)dQ/dt + e^(-t^-1- t^2)(1/t^2 - 2t) Q.
d/dt (e^-(t^-1- t^2)Q) = 0 implues that e^(-t^-1- t^2)Q = c, where c is an arbitrary constant.
Therefore, the general solution is Q(t) = ce^(t^-1 + t^2)

[uncut mason]

correct any errors if you see any

i've assumed that you are aware of what the integration factor is and where it comes from

[blazing zephyr]

bro if the channel already closed you probably should stop helping them cause their inquiry probably was solved

[cedar kiln]

@uncut mason Has your question been resolved?

[uncut mason]

the problem was not resolved

[blazing zephyr]

they closed it

probably meant that they solved it

[cedar kiln]

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[dreamy summit]

Hey there, this might be a silly question, but im new on maths.

Can i solve this function equation for f(x) = 0. Without using the quadratic formula?

[prisma shard]

（x+3)(x-4)

[vagrant elbow]

This is not what we refer to as a functonal equation btw

[sacred grail]

function equation

[vagrant elbow]

And you have 2 ways to do it without the formula, either complete the square, or factorize it

[dreamy summit]

sorry, whats the name?

[vagrant elbow]

Nothing really, solving a quadratic

Or finding its roots

[dreamy summit]

oh, okay

[vagrant elbow]

Have you heard of/done these before?

[dreamy summit]

you refer factorizing it like this

[vagrant elbow]

yes like that

[dreamy summit]

yeah, i heard about factorization

[vagrant elbow]

Right, do you know how to proceed?

[dreamy summit]

mmm, more or less

[vagrant elbow]

Okay, what's bothering you then?

[dreamy summit]

i didnt remember that i can factorize the quadratic😅

[vagrant elbow]

Ah, alright

[dreamy summit]

It's been a while since I do anything related to maths

[vagrant elbow]

Always check for factorization first

[dreamy summit]

Okay

thank you so much

.close

[cedar kiln]

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[cedar kiln]

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[spark stratus]

can someone explain this step to me?
how is the numerator transformed

[austere pawn]

where is the fraction

[spark stratus]

this Fₒ/m [....] is numerator. can't you see this yellow line

[random badge]

Hi @spark stratus
We used sin(A-B) property = sinA CosB - CosASinB

Like R is common in both terms in numerator

We are left with { cos(phi)sin(wt) - cos(wt)sin(phi) } R

[cedar kiln]

@spark stratus Has your question been resolved?

[merry rivet]

rcosxsiny - rsnixcosy = rsin(x+y)

[cedar kiln]

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[long mauve]

Find the area of an isosceles triangle where the base angle is 50 degrees and the length of the leg is 2 units more than the height to the base.
(translated question so if something doesn't make sense lmk)

[long mauve]

leg 💀

[steel heart]

what does leg mean?

[long mauve]

these