help-26

[topaz sinew]

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[quartz mulch]

Help

[quartz mulch]

i have a graph, 5 vertecies and 4 edges, how many trees can i make

I tired finding the number of chained ones i can make, so 4!, + (4!-3!) + (4!-2x3!) + (4!-3x3!) + (0) = 60

then i found the number of ones where one of the vertex would have 3 edges going into it (4C3) * 5 = 20

then 4 edges in each vertex which is a star tree graph, and there is only 5

which accounts for 85 of the tree graphs

how do i find the other 40 and which ones am i missing

[quartz mulch]

reasoning: Found all where starts at the first point (A) i.e., ABCDE, ABCED, etc... ending up with 4! for all ones starting with A, but since none can end in A as we have all ones starting with a and reverses dont count, then ones starting with B and not ending with a = 4! - 3! as there are 3 to choose after B, 2 to chose after, 1 to choose after, then A to choose hence -3!

[quartz mulch]

then for this one we start with A, it picks 3 points and has 1 left to choose from, so theres just 4 points and it has to pick 3 then the last one is connected, so 4C3 (4) x5 (for the other points) = 20

[quartz mulch]

then where all points are connected to 1 point, there are just 5 ways

since a tree has to have (in this case) 5 verticies and 4 edges, what other ways could there be

<@&286206848099549185>

[topaz sinew]

@quartz mulch Has your question been resolved?

[quartz mulch]

<@&286206848099549185> where we at

[quartz mulch]

thanks for nothing, if you can help dm me

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[crisp lodge]

how did we get 7pi/10?

[crisp lodge]

if cos result wasn't negative, it'd be pi/10 and we could pi-p/10 but the answer is 7pi/10, so what gives

[topaz sinew]

@crisp lodge Has your question been resolved?

[opal vault]

you mean if cos wasn't negative it would be 3pi/10?

[crisp lodge]

i mean the sqrt10+2sqrt(5))/4

oh wait shoot you're right

and then you just pi-3pi/10 which gives you 7pi/10

[crisp lodge]

thank you we were stuck on this for ages

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[blissful mantle]

Can someone check if i found the inverse matrix correctly?

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[blissful mantle]

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[jaunty talon]

is this correct?

[smoky sparrow]

how did you get that?

(so it's easier for us to check)

oh wait it's just that

[jaunty talon]

0.80×20=16 liters of water

0.95×x liters of water

Total water

Total water=16+0.95x

[smoky sparrow]

that's correct

[jaunty talon]

to show salt would it be this-

16+0.05x?

[smoky sparrow]

no, cause it's now 20 * 0.20 + x * 0.05

[jaunty talon]

ah so 4+0.05x

[smoky sparrow]

yeah

and the total water + total salt = 20 + x

[jaunty talon]

last question, is this correct

[smoky sparrow]

yep that's correct

try using * for multiply though

[jaunty talon]

ty

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[elfin raven]

what is the shortest distance between a point P<2, 1, 5> and a line L = <1, 2, -5> + lambda * <6, 3, -4>

[elfin raven]

I tried doing it just from using properties and no formula, but I was off

answer was 7, and i got 7.03

[worthy storm]

explain how you got 7.03

i get exactly 7

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[elfin seal]

The frustum of the regular quadrilateral pyramid ABCDA'B'C'D' has AB = 7sqrt2 cm, A'B' = 4√2 cm, AA'=3sqrt10 cm. We denote 0 and O' the centers of the bases. Compute OM, where M is on OO', so that triangle C 'MC is isosceles from the point M

[elfin seal]

I just need an idea to start on

[topaz sinew]

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[fleet fog]

hey just a quick question is there a reason why i’m splitting the equation into two fractions? i don’t understand why and got this right from sheer luck

[fleet fog]

the ratio identities part i understand but i honestly never would’ve known to split it into two fractions if that makes sense

[topaz sinew]

@fleet fog Has your question been resolved?

[polar torrent]

no theres nothing about this that "screams to me" to separate the right hand side. i would be more inclined to write the left side as sines and cosines and find a common denominator

ie. tan = sin/cos and cot = cos/sin and see where that goes

[fleet fog]

hmm ok i thought that was weird seeing it split up into two fractions

[polar torrent]

the only thing that MIGHT make me think to split the fraction is that the denominator has only one term

for instance, you might rewrite (x^2 - y^2)/xy as x^2/xy - y^2/xy = x/y - y/x

[fleet fog]

ahh i didn’t know that

or maybe i did and i forgot

[polar torrent]

lol either way, it's not obvious to me unless i had a leading question to make me think that way

im more likely to use the strategy of common denominators after changing everything to sin and cos

[fleet fog]

yeah i was just confused seeing it solved that way just wanted to know if i was in the wrong doing it another way

[polar torrent]

no! trig equations can often be solved in multiple ways. using an online platform kinda sucks in this sense bc it doesnt give you the flexibility to solve it in a way that makes sense to you

[fleet fog]

yeah i was struggling a bit the way its having me answer these questions is pretty dumb

anyways thank you for helping me out on this

have a good rest of your day

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[topaz sinew]

@ashen spade Has your question been resolved?

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[cosmic storm]

Can this be simplified into a monomial? The instructions say that I have to 😭 but idk how or i don’t even know if it’s possible help

[eager compass]

not possible imo

[cosmic storm]

so my solution is the simplest form already?

[eager compass]

let me check but i think yes

Maybe i have the wrong definition of monomial but it makes no sense to ask for a monomial since here you have trigonometric functions

[cosmic storm]

😢 it says “simplify the expression as a monomial in terms of sinx and cosx”

[eager compass]

oh ok

then write 1/cos(x) + sin(x) and it should be right i guess

Do you need to make it as a monom? bcause if they only want cos or sin this is not possible

[cosmic storm]

yeah I guess there’s a flaw in this assessment (it’s a common occurrence in our school) ill disregard the monomial part and just set my answer as 1/cos(x) + sin(x)

Thank you 🙏 😔

[eager compass]

no problem

happens to the best to make an error in the exercise

[cosmic storm]

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[hollow flax]

Hello, I need some help with this problem. I legit have no idea where to start with this one

[topaz sinew]

@hollow flax Has your question been resolved?

[shy pivot]

I dont have the answer but you could start by saying Z/pZ is a field right ?

Whats the end of the line

Oh mod p ok

If that helps you

[hollow flax]

uhh

idk about

group theory yet

this is a number theory class

[shy pivot]

Oh ok

You could try and prove that there are only 2 solutions then that those they give you are solutions@hollow flax

[hollow flax]

ohhh

hmm i might try that?

i think theres something special abt

[shy pivot]

Then id use wilsons theorem but that might not be in your course

[hollow flax]

the (p-1)/2 term

since it shows up in other problems too

[hollow flax]

yeah

[shy pivot]

Hold on

[hollow flax]

the problem is under the wilsons section

[shy pivot]

Which step are you proving first ?

[shy pivot]

K then it seems better to use it at some point haha

[hollow flax]

yep

[hollow flax]

wdym

[shy pivot]

That only 2 solutions exist

Or that those given are indeed solutions

[hollow flax]

uhh

i actually dont know how to prove the first part

or is it just

[shy pivot]

Ok

[hollow flax]

suppose two solutions exist, then they must equal +-(solution)

[shy pivot]

Give yourself A a solution

[shy pivot]

Not really, that works with reals because x²=1 only has two solutions

In Z/pZ its different

[shy pivot]

Then B another solution

Write a relation between the two

[shy pivot]

Sorry

I mean ifvits i your course sure

Where i am youd have to prove it

[hollow flax]

wait how does this work

i think thats outside the scope

[shy pivot]

No youre correct

Yes youre right but essentially it works because its a field so x²=1 only has 2 solutions

My mistake

You were right, so now we know there are at most 2 solutions

Sorry if i was confusing

[hollow flax]

thats fine

but essentially

i can start with the solutions instead

and show that they are solutions, and unique

right?

[shy pivot]

Exactlt

You know there are at most 2 solutions, so if you find two there are no other solutions

Tell me if you are stuck

[hollow flax]

alright, ill take a break from studying anyways hahaha

thanks for the help

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[proven cradle]

I'm new to trigonometry and I don't understand how to solve both questions

[neon iron]

,rotate

[dense lily]

Are these two questions?

[proven cradle]

yea

[dense lily]

Let's do the first one.

[proven cradle]

i understand the first question but what does tan^2 315° mean?

[dense lily]

(Tan 315)²

,w tan²315

Sad

[dense lily]

@proven cradle

[proven cradle]

oh

does it meant like (tan315)^2 = (-1)^2 or somth?

[dense lily]

Are you talking about the denominator?

[proven cradle]

uh yea?

[dense lily]

Do you know the identity 1+$tan^2\theta = sec^2 \theta$

[thorny flame]

David

[dense lily]

Not really.
It just says
1+ $tan^2 315$

[thorny flame]

David

[proven cradle]

oh

[proven cradle]

so it's sec^2 315?

sorry it really confuses me, english isnt my first language

[dense lily]

Yes.

[proven cradle]

alrr

[dense lily]

No problem

[proven cradle]

now i understand the first question but how about the second one?

[odd forge]

draw a right angled triangle

[dense lily]

P y t h a g o r a s too

[odd forge]

follow this

[proven cradle]

oh

nvm i get it now

ty!!

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[mortal steeple]

<@&268886789983436800> again

[young lagoon]

<@&268886789983436800>

[pseudo sonnet]

.solved

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[mortal steeple]

.close

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[old path]

Not sure why this is wrong (ik it's not bc of not adding stuff in the parenthesis, cengage takes unsimplified answers)

<@&286206848099549185>

[ruby vale]

i'm not familiar with this, is this sneed's quadrant formulae ??

[old path]

Just integrating volume with disk and washer method

[ruby vale]

i see

[old path]

Im not sure where I went wrong in my work lol

Im guessing it's the integral I wrote from the start

But I don't see what the problme is

This is the 2d graph

Resending question

<@&286206848099549185>

[ruby vale]

@old path

does this bare any relevance to your problem?

[old path]

Bro I'm actually trying to get help

[topaz sinew]

@old path Has your question been resolved?

[old path]

<@&286206848099549185> Pls if anyone knows calculus this is the last question I just can't get right and idk why

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@old path Has your question been resolved?

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[midnight sinew]

the proof seems to say that the elements of pRp are of the form x(r/s) where x is in p and r/s is in Rp

but arent ideals generated by a set always written in terms of sums?

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[cinder fractal]

I need to find out how many 3-distinct-digit even numbers there are. My first idea was to multiply 5 (units) * 9 (tens) * 7 (hundreds). Why doesn't that work?

[shut vigil]

There are 10 symbol numbers

0 - 9

so first you can choose 10 numbers

Once you picked a number, the next choice would have 1 less number

so 9 numbers to choose from

then the next choice would have 1 less number to choose from

so 8 numbers onlu

there are 10 choices, 9 choices for the second pick, and 8 choices for third pick

10 x 9 x 8

Ah wait

It saays even but the principle is just the saame

there are 5 even numbers 0 2 4 6 8

so 5 x 4 x 3

or 60 numbers to choose from

but the principle is just teh same as I have earlier

[loud oasis]

but the 10s and 100s place don't have to be even

[shut vigil]

ah fuck yeah righ

so 5 x 9 x 8?

AH WAIT

I GOT IT

its 10 * 9 * 5

You can still pick the even number that was picked earlier

If I got 2 for the ones number, doesn't mean I am not allowed to pick 2 again but for other numbers

Like 222

[cinder fractal]

All digits have to be distinct

[shut vigil]

Ah yeah

[cinder fractal]

My reasoning was all even digits for units (5), all digits except the previous one for tens (9), all digits except the previous two and except for zero for hundreds (7)

But my textbook's answer is 328

[loud oasis]

well the number of digits available for hundreds depends on whether 0 was used in the units or 10s place

[cinder fractal]

ugh

that's way more complicated than i thought

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[short inlet]

Me forgor how to do

[shut vigil]

since x^4 is just (x^2)^2 let x^2 just be a variable

do teh same thing for all

[short inlet]

?

[fair thorn]

substitute t = x^2

factorize

[short inlet]

I’m lost-

[quartz shoal]

Basically what they're saying is, since it's x^4 and x^2, you can use another variable, let's say t, to replace it. Let t=x^2, then the function turns into

3t^2+26t-40

[short inlet]

Which problem is this for- 21?

[quartz shoal]

16

[short inlet]

Oh sorry I was looking at the wrong area of the paper

So factor out x^2…?

[quartz shoal]

Not factoring it out, just substituting in a separate variable. In all honesty though you can also just keep it as x^4 and x^2, because the process doesn't change at all

[short inlet]

Do I use the parts of 40 to add to 26..?

[quartz shoal]

You'll want to multiply by the A term first, so multiple -40×3

[short inlet]

Ohhh yea

On both it just the 40?

[quartz shoal]

Which is -120. Then find the factor pair that adds to 26

[short inlet]

Ok!

Once I get that factor pair..?

[quartz shoal]

What'd you get

[short inlet]

20 and 6 🧍🏻

[quartz shoal]

So there's a problem with that choice. Remmebr that it's -120, so you want a positive and negative number. If either of those are negative it won't add to positive 26

[short inlet]

Fuck

4 and 30 then

[quartz shoal]

Correct, which one should be neg

[short inlet]

4

[quartz shoal]

Great

[short inlet]

Don’t I need to multiply the factors by 3 or smth..?

[quartz shoal]

So now what you're doing is rewriting the middle term with those two numbers. In this case it would be

3x^4+30x^2-4x^2-40

[short inlet]

Huh-

[quartz shoal]

You do not since we've basically already did that

[short inlet]

I just need to factor this-

[quartz shoal]

Correct, that's what we are doing. Unless your teacher does it a different way

[short inlet]

I don’t think this is how she taught us

[quartz shoal]

You'll have to provide how she taught you then so I can use that method

[short inlet]

I’m not sure if I have it in my notes as I lost my original notebook a while back

[quartz shoal]

Unfortunately then I can only use the method im using to help you with this

[short inlet]

I don’t think it matters to her as long as I can do it

[quartz shoal]

Alright

[quartz shoal]

Does this make sense to you then

[short inlet]

Mhm

I wrote it

[quartz shoal]

Ok

So now think of this expression as two separate halves. Like this

3x^4+30x^2 | -4x^2-40

And what you do is factor completely each side separately

[short inlet]

So factor by grouping..?

[quartz shoal]

Yes

[short inlet]

Ok

[quartz shoal]

Let me know what you get

[short inlet]

(3x^2 -4) (x^2 + 10)

[quartz shoal]

Looks great

And that's it

[short inlet]

That’s all..?

[quartz shoal]

Yep, that's the expression factored

[quartz shoal]

If you expand this, you'll get the exact same expression back from the question

[short inlet]

Ok

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[lament fractal]

Does this look correct so far

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[autumn relic]

If a water bucket filled with water spins around in a vertical circular motion around a center, why at a certain speed would the water not fall to the ground in the highest point. (When it is pointing downwards.)

[winter egret]

Note that, on average, the force of gravity shoves the water equally in all directions because the water is rotating

So if the water spins fast enough then we observe the water staying in place

[floral cedar]

what

gravity always pulls down

[winter egret]

from the reference frame of the rotating water gravity appears to be rotating

[floral cedar]

ok

[winter egret]

and thus the net force on the reference frame over time is zero

I'm probably abusing terminology here because I'm not a physicist

of course there are also forces you need to consider from the water hitting the edges of the bucket

but there's probably some kind of similar symmetry argument there

also this server isn't really a physics server

surface tension might also play a role

it's much harder to get water to spill than it is to push water down if it's already spilling

I mean, there is a net force acting on the water obviously because it has acceleration pointing towards the center of rotation

but this net force is generated by the edge of the bucket I think

There is asymmetry within the bucket, one edge pulls the water and the other edge doesn't really do much

and since the direction of this pulling force changes, these forces don't cancel each other out over time unlike gravity

And in order for this "forces cancel each other out over time therefore they're basically zero for all points in time" effect to be noticeable you need the system to evolve very quickly in time (i.e. the bucket has to be spinning quickly. If you stop the rotation in mid-air then the water is just gonna fall out and the system will have a different composition even if you symmetrically also stop the rotation for the same amount of time in the resting state)

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[timber gorge]

The question: Analyze and sketch a graph of the function. Identify any intercepts, asymptotes, relative extrema and points of inflection.

[timber gorge]

I don’t know how to get the derivative to figure out the rest. Also I need help graphing the function.

This is Calc

[inner crow]

Do you know quotient rule?

[timber gorge]

Yes

[inner crow]

Then you can use that on the function to get the derivative

[timber gorge]

Ok I’ll try

I tried to do the quotient rule but I am confused now and don’t know what to do

Wait I did something wrong sorry

[inner crow]

You didn't need to factorise the top, that's just overcomplicating it for yourself

Anyway, what was it you needed the derivative for

Oh, for stationary points. Just make it = to 0, solve the equation for x, then you can do a 2nd differential to find out which one's max, min, and POI

[timber gorge]

I needed to figure out the relative max and min. I also needed to figure out the POI which I need second derivative and the chart

[inner crow]

Hint: It should make sense that a fraction will only ever be = to 0 if the numerator is = to 0

So, assuming your derivative is right, haven't checked it, but if you're confident in it, then what are your x values for when x^2 + 10x + 8 = 0?

Then to get the actual coordinate of the points, you put in your calculated x values into original equation and get y values

Then, to find which ones are min, max, or POI, you do 2nd derivative using quotient again

[timber gorge]

I’m still stuck on finding the first derivative and I’m confused if I’m supposed to simplify it all the way down to this

Do I make it equal to 0 and solve for x?

[inner crow]

Okay, so, let's say u = x^2 + 4x + 4

what is u'

Then do the same for v = x+1

[inner crow]

You can't cancel the x+1 because the numerator is a -, you'd need to cancel the x^2 + 4x + 4 as well if you did that

[timber gorge]

So what should I do? I’m sorry, I didn’t understand the lesson my teacher taught me.

[inner crow]

u' means differential

So, what is the derivative of u

[timber gorge]

2x+4

[inner crow]

Yep

and v' is?

[timber gorge]

1

[inner crow]

Yep, so u = x^2 + 4x + 4
u' = 2x + 4

v = x + 1
v' = 1

[timber gorge]

Yea

[inner crow]

And then you do what you did at the start, so (vu' - uv')/v^2

And, as you showed, that gives you ((x+1)(2x+4) - (x^2 + 4x + 4))/(x+1)^2

Try expanding out (x+1)(2x+4)

And then simplifying the whole numerator

Tell me what you get for the numerator when you've got that

[timber gorge]

2x^2+6x+4

Wait the whole numerator

Oops

[inner crow]

dw, the part you did is correct, now just include the other part and simplify

[timber gorge]

The numerator is x^2-2x

[inner crow]

Close, just 1 small mistake

How would you write -(x^2 + 4x +4) if you had to get rid of the brackets?

You got it right for some of them, but not all

[inner crow]

Exactly

So what's 6x - 4x

[timber gorge]

2x

[inner crow]

Yep, you see the mistake now?

x^2 + 2x would be the numerator

[timber gorge]

Oh yes sorry

[inner crow]

dw

So now you have (x^2 + 2x)/(x+1)^2, which is much nicer

Missed the fraction for a sec

You don't need to do anything to simplify this

You just set it = to 0

Is that okay?

[timber gorge]

Yes. So the x=0 and x=-2 ?

[inner crow]

yep

As for finding out if it's min, max or POI, you probably know how, just remember to do chain rule when differentiating (x+1)^2, or you can just expand it to x^2 + 2x + 1

[timber gorge]

So I should do the quotient rule again for the 2nd derivative

[topaz sinew]

@timber gorge Has your question been resolved?

[timber gorge]

<@&286206848099549185> I still need help

So did I make a mistake? I’m confused.

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[delicate pulsar]

Given the function 𝑔(𝑥)=((18−2𝑥^2)/(2𝑥^3+5x^2−9𝑥−18)) , simplify the expression, determine the x-intercepts, y-intercept, horizontal asymptotes, vertical asymptotes, limits and graph the function.

[delicate pulsar]

I need help finding the limits and finding simplifying g(x)

[hushed apex]

Are you wondering which x values you must observe for your limits, or are you rather having trouble solving your limits? What have you tried so far?

[delicate pulsar]

as of now, i am getting a very complex equation

[delicate pulsar]

i have issues with solving and going through with my limits

[hushed apex]

Ok, let's try one of the limits together. Which one would you like to do?

[delicate pulsar]

i would like to start off with x approaches -inf., y approaches...

[hushed apex]

Sure, so when you try and solve the limit, do you get an indeterminate form? Which indeterminate form do you get? Also feel free to ping me

[delicate pulsar]

to be quite honest, i always struggle with finding limits, which causes mayhem, especially when I go through with graphing my rational graphs

ALSO, if possible, could you walk me through the stesp from factoring to graphing?

I understand how to factor the equation, although, I am getting VERY complex numbers (sqrts), which should not occur, as the work we were given shows the answers, which are ONLY common factors such as (x-2), (x-3), (x+4) & such!

@hushed apex

here is an image of everything that should be done:

[hushed apex]

Sure we can factor out the numerator and the denominator of g.
We can start by the numerator : 18 - 2x^2.

When you want to factor something, always look for a GCF (greatest common factor) between all terms of your polynomial.

Do you see a GCF in this expression?

[delicate pulsar]

2?

∴ (9-x^2)

[hushed apex]

Almost

[delicate pulsar]

∴2(9-x^2)

[hushed apex]

Yes (also, don't use the "therefore" symbol here. Try to use the biconditional symbol (⟺))

Now, by convention, we always want the x term with the highest exponent to be positive

So we would also factor out a "-" here

[delicate pulsar]

right! so it will be -2(-9+x^2) OR -2(x^2-9)

-2(x^2-9)

[hushed apex]

Precisely, yes. Now, there is a factorization technique you can use for this binomial (x^2 - 9). Please tell me which factorization techniques for binomials do you know of, and tell me which one do you think we should use.

[delicate pulsar]

difference of squares?

(x-3)(x+3)?

QUICK QUESTION, off topic, but can we voice chat?

[hushed apex]

No unfortunately

Sorry

[hushed apex]

But yes this is correct

[delicate pulsar]

all good

ok sounds goods

[hushed apex]

Just don't forget the factor

the GCF

[delicate pulsar]

ok thanks!

[hushed apex]

np!

[delicate pulsar]

now how would I find the factors of the denominator?

[hushed apex]

For 2𝑥^3+5x^2−9𝑥−18 it's a bit harder.

As usual, we first try to see if there is a GCF. Is there a GCF here?

[delicate pulsar]

nope

[hushed apex]

Now, the way I learned how to solve these is pretty strange, so maybe your teacher wants something else.

First, we have to find one of the y-intercepts of 2𝑥^3+5x^2−9𝑥−18 via trial and error. In other words, we want to find a value of x such that if we substitute said value of x in that expression, we get 0. Try and find one via trial and error!

[delicate pulsar]

so do i use integral zero theorem?

and get a quadratic, which i later factor?

[hushed apex]

Woah never knew that had a name 😳

[delicate pulsar]

LOL

I THINK THATS WHAT IT IS

one secondd

like this?

here!

or by using synthetic division???

@hushed apex

[hushed apex]

Seems like you're getting the right answer so it's fine ig

Try it with the other polynomial now

[delicate pulsar]

ok

i got:

g(x)=(x-2)(2x^2+9x+9)

which has to be further factored

so I am a little embarrassed to say this, but I tend to always forget how to factor quadratics

@hushed apex

@hushed apex ?

i am trying to find my chat

on my computer, but cannot find it

[hushed apex]

Sorry, I'm here. There are two ways to factor quadratic equations. The first way is to use the quadratic formula, the second is to find a pair of two numbers (say, i and j) which satisfy the following two conditions :

• i * j = a * c
• i + j = b

where a, b and c represent the coefficient of a quadratic equation (ax^2 + bx + c)

[hushed apex]

@delicate pulsar this?

[delicate pulsar]

yes

my user is not there, or my help chat for #26

never mind FOUND IT

ok so for this

is it 6 & 3?

but then if i make an equation, wouldn’t it be (2x+6)( x+3)

which is wrong @hushed apex

[hushed apex]

   2x^2 + 6x + 3x + 9
⟺ 2x(x + 3) + 3(x + 3)
⟺ (2x + 3)(x + 3)```

You most likely made a mistake when factoring at the end

[delicate pulsar]

oh right!

i got it!!

ok so now my final factored equation is?

[hushed apex]

Let's try to figure out. We know that g(x) = (x-2)(2x^2+9x+9).
However, we also know now that 2x^2+9x+9 = (2x + 3)(x + 3)
What can we conclude?

[delicate pulsar]

g(x)=((-2(x+3))/(x-2)(2x+3)), x not= 2

not=3

[hushed apex]

Ahh wait no sorry

[hushed apex]

This is incorrect. g(x) is not equal to this. This is only the denominator of g(x).

[delicate pulsar]

waht does that eman?

i got this

hope you can read my writing?!

[hushed apex]

There is a mistake here

Try to find it!

(Technically there are two mistakes, but yeah)

[delicate pulsar]

the numerator should have (x-3)

and the end bracket for the denominator?

[hushed apex]

Instead of x + 3

[hushed apex]

yes

[delicate pulsar]

my faulty

im making so much small mistakes which cause issues later

thanks so much for that!

so after this, i should be able to find the x & y

i’ll do that and send it here

[hushed apex]

Find the x and y? wdym?

[delicate pulsar]

like teh intercepts

so that i could find the graph

so the intercepts, the verical and horizontal asumptotes and then the limits, and graphing it

no worries

see you, thanks for your help!

can i continue with another person?

<@&286206848099549185>

.close

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[solar chasm]

I'm currently working on Problem 5 and it is a Sturm Liouville Eigenvalue Problem. I have found the characteristic equation and finished the other two cases but am stuck on the case where lambda is a complex root. I have simplified it all that I can and am not sure how to solve for lambda. I know that it will be in the form (lambda)_n, but unsure how to get rid of the sine and cosine.

There was a previous problem in the homework similar but it only had sine, not sine and cosine.

The two eigenvalues and eigenfunctions are given, (the professor gives the answer and we just have to get to it so we know what to expect to get) and we have to use the fsolve command in matlab to find the first two eigenvalues.

I just to know how to isolate and solve for lambda from the last equation of the last picture.

Problem 6 has a similar outcome except the sine and cosine are interchanged and one is negative

[solar chasm]

My bad. Didn't see that it was for pre-uni questions only mainly.

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[river harbor]

hello, I am struggling with this question on my assignment

topic is discrete mathematics

my work for a) so far is the following:

prove p->q using direct proof
prove q->p using contrapositive

p->q worked but when i did contrapositive q->p (so -p -> -q)
i get the statement: if n even, then n(n+1) is odd

plugging n=2k in n(n+1) i obtained 2(2k^2 + k) which is even

if it doesnt follow the contrapositive, then I have not successfully proved q->p right?
which means i have proved the statement to be false ?

[topaz sinew]

@river harbor Has your question been resolved?

[river harbor]

<@&286206848099549185>

i tried doing c) aswell and obtained the following
true by direct proof for p->q
true by contrapositive for q->p

i dont know how to verify b) though

nevermind, i proved b to be false with an easy counter-example

number = sqrt(2) (which is a known irrational number)
(sqrt(2))^2 = (2) which is a known rational number

direct proof failed

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[neon iron]

can someone please tell me basically what is this problem called?

basically what do I look up to learn how to do function problems that include square root

[heady robin]

If you could specify more about what’s your objective exactly

You want to learn how to deal with equations that have root in them?

[neon iron]

specifically this type of equation which is a function

i am taking precalc

basically I have no idea how he got these answers

i need to learn from scratch how to solve a f() = sqrt problem

[heady robin]

Not really

You know how to solve normal equations right?

By reverse engineering the pemdas method right?

[neon iron]

nope

what do you mean by normal equations?

[heady robin]

Like 3x +2 = 12-2x

[neon iron]

hmm

yes

i have to get xs on the left regulars on the right

so it would be 5x = 10

to 2

[heady robin]

Right

More specifically im asking about which opreation do you start with

[neon iron]

i start by adding 2x

[heady robin]

And subtracting 2 right?

[neon iron]

yea

[heady robin]

Then you for the final step you divided by 5 right?

[neon iron]

yea

[heady robin]

You went like this:

1)Addition & subtraction

2. multiplication & division

You didn’t need multiplication this time

[neon iron]

this time i needed to cancel the square root by doin 0.21^87

???

[heady robin]

No

[neon iron]

ok shit

[heady robin]

What comes after these two is exponentials

Just like pedmas but in reverse

[neon iron]

right....

[heady robin]

P = parentheses

E= exponentials

M= multiplication

D= division

A = addition

S = subtraction

[heady robin]

In short the square roots come in the exponentials step, and you cancel them by raising each side by power of 2

Sqrt of x = 4

(Sqrt x)^2 = 4^2

X = 16

You got my point?

[neon iron]

yea but

where did you get ^2

so you would just do the sqrt of 4

which is 2 then you ^2

[heady robin]

When you write the sqrt

[neon iron]

?

[heady robin]

What do you put in the index place

[neon iron]

i dont know wha the index place is

x?

[neon iron]

oh so its always 2

had no clue bro

[heady robin]

Yeah

[heady robin]

Now you do

[neon iron]

𝑓(0) = 87√0 + .21

so here

[heady robin]

If it was 3 you would do ^3 for 4 you do ^4 and so on

[neon iron]

wait if what is 3?

i though it was always 2

[heady robin]

You know what f(0) mean right?

[neon iron]

yea f is the input

you just replace all instances of x with 0

[heady robin]

If not shown it’s 2, unless specified otherwise

If it’s not 2 it will be shown

[neon iron]

so in this case its 87

so I would do 0.21^87

no thats not right

[heady robin]

You will see smth like this

3 would change to any number like 4 5 6

But if not shown it’s always 2

[neon iron]

so to focus on this first question

[heady robin]

Correct

Yeah

[neon iron]

what exactly is 87

its not ^87

like you said its not a 2 or a 3 up there

its just to the left

[heady robin]

87 * sqrt(0+0.21)

[neon iron]

oh ok times....ok now i see....

its literally like 87(sqrt...)

[heady robin]

If no sign is there, it’s usually multiplication

[neon iron]

lol

[heady robin]

Yup

That’s why he divides by 87 in the third question

[neon iron]

so i do sqrt of 0.21

then times 87?

[heady robin]

Yeah

[neon iron]

oh shit

[heady robin]

Use calc

[neon iron]

it actually is the answer

[heady robin]

Yup

[neon iron]

wtf thats so easy bro haha

i think I can figure out the rest

thanks man

[heady robin]

Np

[neon iron]

🙏

[heady robin]

You know why and how he changed from mins to hours right?

In 2nd question

[neon iron]

nope

not exactly the reason why

easier to display as a decimal or something

oh t is in hours

[heady robin]

Exactly

[neon iron]

actually glad i didnt leave

i dont understand how he does the second question

when i plug in f(45) it is 607.etc

[heady robin]

45 is minutes

[neon iron]

oh i see

45 is 0.75 out of an hour

[heady robin]

Change it to hours for it to be valid input

[heady robin]

Yup

Don’t send .close till you get everything correct and understood

Any problem, just ping me

[neon iron]

alright fasho thanks again bro

[heady robin]

Anytime

[neon iron]

.close

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[royal sky]

why (-1)^3.5 on calculator for example

[royal sky]

is error

[loud oasis]

(-1)^3.5 = (-1)^3 * (-1)^0.5

[royal sky]

and (-1)^0.5 is error?

so anything not whole nyumber as exponent

is

error

[loud oasis]

well it's the imaginary number i, which would throw an error if your calculator only supports real numbers

[royal sky]

when its (-1)

[civic nymph]

yes because $-1^{0.5}=-1^\frac{1}{2}=\sqrt{-1}$

[royal sky]

ohhhhhhhhhhhhhhh

yeah

[thorny flame]

Joshii

[royal sky]

okay ty

[civic nymph]

oops i havent used this in a while

[royal sky]

.close

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[torpid shard]

given two subsequent numbers on a Padovan Sequence, \$f(n) = f(n-2) + f(n-3)$
is it possible to know exactly the two previous number to those chosen?

[thorny flame]

zzz0nnn

[torpid shard]

This could somewhat useful for another bigger problem of mine, but which doesnt necesarily correlate to the problem itself.

Still, im not mentally able to tackle the problem, ik that with 3 numbers is possible, but idk about 2

i figured that no, its not possible, since padovan sequences require to define 3 initial numbers

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[cinder rapids]

for part a)

if we substitute in (0, 4-2e), B = 4

but e^0 is 1

so that would mean A = -2e? but the answers say that A = -2 which isn't possible unless e^kx = e which isnt possible because x = 0

am i wrong or is the question wrong?

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@cinder rapids Has your question been resolved?

[cinder rapids]

<@&286206848099549185>

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[ivory sorrel]

I think in order to get an idea for denseness, it's better I start by proving that the set $\N$ doesn't have an upper bound in $\R$

[thorny flame]

Mr bean is not $\R \setminus \Q$

[ivory sorrel]

My definition is if $A \subseteq \R$ and $B \subseteq \R$, such that $x \in A ; y \in B \implies x \leq y \forall x \in A , y \in B $. Then there exists $\alpha \in \R$, such that $a \leq \alpha \leq b$

[thorny flame]

Mr bean is not $\R \setminus \Q$

[ivory sorrel]

So to start, let $n$ be the largest natural number. so $A= {1,2,3 \dots, n}$ And let $B$ be the set of all real numbers greater than or equal to $n$. From this it follows that $n-1 \in A$ as the naturals are closed under subtraction for all numbere greater that or equal to 2. Let $a$ be the least upper bound of $A$. So $n-1 \leq a \implies n \leq a+1$.

What do I do now?

[odd pagoda]

the naturals are not closed under subtraction

[thorny flame]

Mr bean is not $\R \setminus \Q$

[odd pagoda]

I really dont know where you are trying to go

clearly n<=a because n in A and a is an upper bound of A

and clearly n+1 is a natural number

and nothing of this has talked about real numbers yet

[ivory sorrel]

Let me try again , let their exist an element $M \in \R$, that acts as an upper bound for $\N$. Let $n$ be this largest element of $\N$ . It thus follows that $ n \leq M$. It also follows that $n-1 \leq M$

[thorny flame]

Mr bean is not $\R \setminus \Q$

[odd pagoda]

there is no largest element of N

[ivory sorrel]

I know

[odd pagoda]

whether there is a largest element of N is separate from the question whether there is an upper bound

there also is no smallest element of {x in R: x>0}. but that set still has a lower bound

[ivory sorrel]

Mhm

Yeah

[odd pagoda]

what are you allowed to use about R

[ivory sorrel]

Completeness axiom, and all the basic properties of R

Like inequalities

[odd pagoda]

"all the basic properties except for the basic property that N is unbounded" ?

what else not

[ivory sorrel]

Okay, the inequality properties

And equality properties

[odd pagoda]

what inequality properties are you talking about

[ivory sorrel]

Like x>y implies x+1>y

[odd pagoda]

ok

do you have floor/ceil

[ivory sorrel]

I suppose so

[odd pagoda]

then use those

[ivory sorrel]

That would just prove there is no upper bound in $\N$

[thorny flame]

Mr bean is not $\R \setminus \Q$

[odd pagoda]

would it?

[ivory sorrel]

yes

[odd pagoda]

how

[ivory sorrel]

The floor function and cei functions map to $\N$

[thorny flame]

Mr bean is not $\R \setminus \Q$

[ivory sorrel]

oops, I mean $\Z$

[thorny flame]

Mr bean is not $\R \setminus \Q$

[ivory sorrel]

Let $A$ be the set of naturals, so $A={1,2, \dots, n}$ Let $B$ be the set of all upper bounds . By the completeness axiom , there exists a lest upper bound $\alpha$.So $n \leq \alpha$. Which means that $n+1 \leq \alpha $. So $n\leq \alpha -1$. But $n\leq \alpha$ , we've thus arrived at a contradiction

[thorny flame]

Mr bean is not $\R \setminus \Q$

[ivory sorrel]

Is this fine instead?

.close

[topaz sinew]

Channel closed

Closed by @ivory sorrel

Use .reopen if this was a mistake.

[odd pagoda]

.reopen

no

[topaz sinew]

✅

[odd pagoda]

A is still not finite

your proof should start with the words "suppose for contradiction that there exists a least upper bound alpha of N. then ..." or something similar

@ivory sorrel

[ivory sorrel]

I see, okay

Let $A$ be the set of naturals, so, $A= \N$ . Let $B$ be the set of all upper bounds .For the sake of contradiction, let , there exists a lest upper bound $\alpha$.So $n \leq \alpha$. Which means that $n+1 \leq \alpha $. So $n\leq \alpha -1$. But we've already established that the least upperbound of $\N$ is $\alpha$. We've thus arrived at a contardiction

[thorny flame]

Mr bean is not $\R \setminus \Q$

[odd pagoda]

for which n does that hold

[ivory sorrel]

All $n\in \N$

[thorny flame]

Mr bean is not $\R \setminus \Q$

[odd pagoda]

ok

[ivory sorrel]

Cool, thanks

.close

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[wheat tree]

It's supposed to be 1/2 what did I do incorrectly?

[wheat tree]

.close

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[balmy roost]

$\lim_{x \to 0^{+}}\frac{1}{x}$

[thorny flame]

Hamdy Hisham

[balmy roost]

Is this limit = $\infty$ or it doesn't exist ?

[thorny flame]

Hamdy Hisham

[tall cove]

It's 2

Yeah the answer is 2

[warm monolith]

saying that the limit goes to infinity is basically just a fancy way to say that it doesn't exist

that is, it grows without bounds

so there's no limit

[balmy roost]

Yup

But what to do in a test

Should I say it's infinity or should I say jt doesn't exist

[haughty mural]

It is infinity

[warm monolith]

since there are different ways in which a limit doesn't exist then you should specify it and say that it goes to infinity

[balmy roost]

Thnx

.close

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[signal stirrup]

is this correct🥲

[topaz sinew]

@signal stirrup Has your question been resolved?

[signal stirrup]

<@&286206848099549185>

[blazing mirage]

.

[signal stirrup]

hi

[silent siren]

"semi-anually" meaning "twice per year"?

[signal stirrup]

yeah

[silent siren]

no

it's wrong

What did you do to get that answer?

[silent siren]

I think you've used anually, instead of semi-anually

[signal stirrup]

so what will I do

.close

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[dreamy lion]

can someone help me with the 5th, the answer is 10cm but idk how

[dreamy lion]

ive tried different approaches I get 13.3,16.67. please help

[topaz sinew]

@dreamy lion Has your question been resolved?

[grave comet]

Try creating a system of equations and use the fact that all 3 angles of the big triangle are 60 degrees, so you pretty much just have smaller equilateral triangles

[dreamy lion]

oh

[dreamy lion]

I found it out, I js had to find the relations

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[thorn surge]

can someone explain me concept of parity

[thorn surge]

how did my teacher get this?

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[glossy girder]

If a geometry problem can be proved via an external construction, is it true that it can also always be proven without the construction, because it seems absurd to be that introducing extra structures somehow adds more information, because it doesn't, so shouldnt we somehow be able to bypass construction in any geometric proof?

[glossy girder]

In other words:
If a geometry problem can be proven using an external construction (such as adding extra lines or shapes), is it always possible to prove the same result without using that construction?

It seems to me that introducing extra structures doesn’t add new information, so shouldn't it be possible to prove the result without them?

<@&286206848099549185>

<@&286206848099549185> ?

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[glossy girder]

.reopen

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✅

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[fresh adder]

to convet tan^-1 to cot^-1 or vice versa we can weite tan^-1x = cot^-1(90-x) or is it 45-x??

[thorny flame]

krypton

[sweet shard]

,w arctan(x) + arccot(x) == pi/2

[thorny flame]

Results provided by WolframAlpha

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Requested by rie.mann

[sweet shard]

i don't understand your question

[fresh adder]

im asking can we write tan^-1x = cot^-1 pie/2 - x

like in sinx = cos90-x

the maths teacher said to do smth like that but i forgot if it was pie/4 or pie/2

[sweet shard]

there are lots of trig identities. be more specific

[fresh adder]

im asking that like sinx = cos90-x there is an identity for tan^-1x= cot^-1 pie/2 - x right?

im asking if its pie/2 -x or pie/4 - x

[fresh adder]

?

[sweet shard]

can't help with something so vague. look there: https://en.wikipedia.org/wiki/Inverse_trigonometric_functions

[fresh adder]

to convert tan^-1x into cot^-1 what would u do?

[sweet shard]

what

if you can't remember more details of the task, i can't help

[fresh adder]

how to slove cot^-1 ( [2tamx]/[1-tan^2x])

we get cot^-1(tan2x)

[sweet shard]

,tex .double angle

[thorny flame]

riemann

[sweet shard]

have you seen the bottom equation before

[fresh adder]

ye

[sweet shard]

compare this

with this

and thiss

[fresh adder]

.

[sweet shard]

with this

right so you got it

[fresh adder]

im asking its cot^-1(tan2x)

to solve further we need to make cot^-1 into tan^-1

[sweet shard]

right and i said it's right

why do you think you can "solve further" ?

[fresh adder]

because my maths teacher said so

-+

[sweet shard]

maybe draw a triangle to express tan(2 x)in terms of cot(stuff)

[fresh adder]

what grade r u in btw?

.close

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[sweet shard]

can't help you if you just want answers

[fresh adder]

i just wanna know how to express cot^-1x in tan^-1 form

😭

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[sweet gale]

Can any one please help me with this question

[burnt badger]

You can use law of cosines to find the missing side length

Then use law of sines to find the Angle at vertex A

Have they shown you those?

[topaz sinew]

@sweet gale Has your question been resolved?

[sweet gale]

How do I use it

Cause there parameters aren't complete

[burnt badger]

If you have a known

Side-Angle-Side

Structure

This is a case for law of cosines

[sweet gale]

Tell me more

Please

[burnt badger]

The triangle ABC

You know the measure of angle B

And you also know the sides that are adjacent to Angle B

This is what i would call a

Side-Angle-Side structure

This type of structure allows you to find the Side opposite from the angle using law of Cosines

In this case, the long side across from Angle B can be found

You can setup law of Cosines

b^2 = a^2 + c^2 - 2ac×Cos(B)

B = angle B

a and c = the two known sides

b = the side opposite from Angle B

You know everything except little b

Once you find the little b

You can then use Law of Sines to find Angle A

Which let you get the bearing

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[radiant jacinth]

ive got 1/x + 1/y * dy/dx = 1 but im not sure where to go further

i ended up with dy/dx = -y/x on the second try but im not sure if thats correct or not

[golden blade]

you didnt do product rule as it seems

but it would have been easier if you did e^(...) on both sides before differentiating

[radiant jacinth]

yeah.. my teacher said we were supposed to use log rule here

[golden blade]

,, \frac{1}{xy} \cdot \left ( y + x \frac{\dd y}{\dd x} \right ) = 1

[thorny flame]

bacc (unhelpful)

[radiant jacinth]

wait okay

so what r u supposed to do next after this

y/xy + x/xy?

oh wait

[golden blade]

well you can solve for dy/dx

or you can plug in x = 1

and the y value, which you get from ln(yx) = x

and then solve for dy/dx

[radiant jacinth]

y + x(dy/dx) = xy??

[golden blade]

to figure the slope

[golden blade]

ya

[radiant jacinth]

oh i see

that would be dy/dx = xy-y over x

[golden blade]

ya

[radiant jacinth]

and then plug in 0

[golden blade]

0?

[radiant jacinth]

1

not 0

brainfart

and then from ln(y(1)) = 1 you get e

i think

[golden blade]

ye

[radiant jacinth]

e-e over 1 would be 0

right

[golden blade]

yes

[radiant jacinth]

bruh

okay tysm

.close

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[sweet shard]

$\log(y) = x - \log(x), \ \frac{dy}{y} = dx - \frac{dx}{x}, \ \frac{dy}{dx} = y-\frac{y}{x}.\ y' = e - e = 0$

just in case you want to use log properties

[radiant jacinth]

ah i see

cool tyy