help-10

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[obtuse pebble]

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[wooden obsidian]

The line y = x+1 intersects the curve y = x^2 +x-3 at 2 points.Find the coordinates of the 2 points

[fathom flicker]

!help

[obtuse pebble]

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[stone forge]

hi

[stone forge]

i just have a question

you know how you conjugate when u wanna verify trig identities sometimes

[zenith raft]

lmfao

[stone forge]

why the hell are u talking here

im using the channel

[fathom flicker]

!occupied

[obtuse pebble]

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

[stone forge]

also wtf is that banner

this generation 🤦‍♂️

[zenith raft]

what generation

[stone forge]

get out

i need help with my trig identities

[zenith raft]

you haven’t even posted a question dawg

[stone forge]

yes i have

[zenith raft]

just body shamed me…

[stone forge]

but ur stupid messages have covered it up

yeah surely i did

get out

[high lily]

you know how you conjugate when u wanna verify trig identities sometimes
yes.
conjugates are relevant in many questions related to verifying identities

[stone forge]

okay my main concern is

im not sure on when you are supposed to conjugate

[high lily]

consider
c^2 = a^2 + b^2
c^2 - a^2 = b^2
(c-a)(c+a) = b^2
conjugates will usually be helpful if you have (c-a) or (c+a) present

[stone forge]

ah i see

wait do you have any examples perchance

in trig identities

[dense imp]

I guess one example I can think of is if you think back to difference of 2 squares: x^2 - y^2 = (x + y)(x - y)
there are some trig identities that involve sums or difference of two squares, like for example
cos^2 x - sin^2 x = cos 2x

maybe you see you have (cos x + sin x) and you think you might wanna transform that for some reason into cos 2x and you multiply both sides by (cos x - sin x) to get the above identity and be able to apply it.

you might also be able to apply stuff like sin^2 x + cos^2 x if you move quantities around from one side to the other of the equation and signs are changing, so i'd say if you are familiar with the couple basic identities that involves differences and squares of sums you can often "smell these out"

[fathom flicker]

you can't just say perchance

[stone forge]

💀

[zenith raft]

why the hell are u talking here

[stone forge]

i had a feeling someone was gonna say smth ab that

[stone forge]

buddy

[dense imp]

!occupied

[obtuse pebble]

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

[stone forge]

can i just say ima conjugate when when it looks like this

[obtuse pebble]

@stone forge Has your question been resolved?

[high lily]

conjuate when a pythagorean trig identity is applicable/relevant

[stone forge]

alr

i think i got it

ty

[obtuse pebble]

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[fluid snow]

Find the inverse of $f(x) = \frac{3x}{2x^2 - 18}$. So we have that \begin{align*} \frac{3x}{2x^2 - 18} &= y \ 3x &= y(2x^2 - 18) \ 3x &= 2y(x + 3)(x - 3)\end{align*} but all this doesn't seem very helpful

[fluid snow]

I also thought about dividing by x right away

But having $\frac{3}{2x - \frac{18}{x}}$ doens't seem very helpful either

[haughty coyote]

Just expand
It's a quadratic in x

[fluid snow]

What do you mean, where?

[haughty coyote]

You wrote 3 lines of computation
You can find where

[fluid snow]

$3x = 2x^2y - 18y$

You must mean that?

Hm

Oh, so you'd just solve for x with the quadratic formula?

[frank dome]

yeah

[fluid snow]

Alright, thank you

.close

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[fluid snow]

.reopen

[obtuse pebble]

✅

[fluid snow]

So I got $\overline f(x) = \frac{3 \pm 3\sqrt{1 + 16x^2}}{4x}$. So here, we don't have to decide which one to pick, the $+$ or $-$ version?

[fluid snow]

Also, the domain of $f$ is $\mathbb R \setminus {3, -3}$ but our functions don't cover $0$, do they

So did we really find the inverse?

[frank dome]

you pick both

it's +-

[frank dome]

wdym?

if you mean the domains dont match, then they arent supposed to. the domain of a function is the range of its inverse function

[fluid snow]

Exactly

But 0 is not in the range

[fluid snow]

We'd have to introduce a second function, right?

A function maps an x-value to a single y-value

Not two

[frank dome]

it is

if you conssider the limit

then f^-1 (0) = 0

[frank dome]

right well i just noticed you mentioned the domain of f is R \ {3, -3} so you would only consider the plus

since the range of f^-1 would then have to be the domain of f

and also if you consider that domain of f then it doesnt have 0 in its range either, and f^-1 doesnt have 0 in its domain either, so it matches

[fluid snow]

,w range (3 + 3sqrt(1 + 16x^2))/(4x)

(-3, 3) is missing

,w range (3 - 3sqrt(1 + 16x^2))/(4x)

[frank dome]

it is supposed to be

[fluid snow]

No, I mean R without the value 3 and without the value -3 with that

[frank dome]

ah

[fluid snow]

It's perfectly fine to have 0 as the argument

It'd be f(0) = 0/-18 = 0

[frank dome]

i thought you were just defining the function to have that domain

[fluid snow]

So 0 should be in the domain of f

[frank dome]

my bad

[fluid snow]

I had to determine the largest domain on which f had an inverse

[frank dome]

well if your original function isnt 1-1, your inverse wouldnt be either, now, would it?

[fluid snow]

And I determined R \ {3, -3}

Maybe that's wrong?

[frank dome]

if you exclude the entire (-3,3) interval then the function is 1-1

but right now it isnt

[fluid snow]

I used the first-derivative method

$f'(x) = -6 \cdot \frac{x^2 + 9}{4(x+3)^2(x-3)^2}$ I got

And that's negative on its entire domain

[frank dome]

decreasing is not the same as 1-1

[fluid snow]

So f is 1-1 there

[fluid snow]

If it's decreasing the whole time it should be 1-1?

Oh wait

The -3 and 3 cause the problems here

It "jumps up"

[frank dome]

it has a discontinuity at x = -3, for example, so it can go back up and decrease from a high value

yes

that's why i thought you excluded the whole interval

since that makes it 1-1

[fluid snow]

How do you know?

[frank dome]

it removes the part where the function "jumps up" and decreases from a high value

so now it just purely decreases the entire time

[fluid snow]

But just looking at the derivative, you can't really tell when it will be at a high value

So you imagined its graph?

[frank dome]

looking at the function, you can

think of it's behaviour just before -3 and just after 3

highly negative before -3 and highly positive after 3

so it jumps from decreasing and negative to decreasing and positive

implying it is 1-1

[fluid snow]

Is there a purely algebraic way?

Maybe using $f(x_1) = f(x_2)$

[frank dome]

ig after you show f is decreasing you can just show that f is negative for x < -3 and positive for x > 3

[fluid snow]

$\frac{3x_1}{2x_1^2 - 18} =\frac{3x_2}{2x_2^2 - 18}$

[frank dome]

this looks unnecessarily painful

[fluid snow]

Ok, thanks

.close

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[cunning flower]

$U_0=2 \ U{n+1}=\frac{1}{U_n}+\frac{U_n}{2}
\text{prove that }\forall \ n \ in \ \mathbb{N} \ U_n>\sqrt{2}}$

[warm shale]

brahim3579
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[cunning flower]

how can i slove it

[solar meadow]

How can there be $U_0$ then?

[warm shale]

Roman_Garland

[cunning flower]

U_0 is the first terme

[solar meadow]

?

[cunning flower]

Natural numbers

its the symbol

<@&286206848099549185>

[solar meadow]

then n cannot be 0?

Bc Natural numbers starts with 1?

does it start with 0 in here?

[cunning flower]

its always start with 0

0 is a decimal number? absolutely not

$U_0=2 \ U{n+1}=\frac{1}{U_n}+\frac{U_n}{2}
\text{prove that }\forall \ n \ in \ \mathbb{N} \ U_n>\sqrt{2}}$
someone can helps here

[warm shale]

brahim3579
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[cunning flower]

<@&286206848099549185>

[solar meadow]

Prove that U_n is always less than sqrt(2) and then is always equal to sqrt(2)

@cunning flower

[obtuse pebble]

@cunning flower Has your question been resolved?

[obtuse pebble]

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[slender lily]

Can someone please help me with this question (it’s translated from German to English and there is a picture which shows the graph needed) Fig. 1 shows the graph of the derivative function f' of a function f. Which of the following statements are correct? Justify your answers.

A: The function f is monotonously falling in the interval [0; 2].

B: The function f has at the point x = - 1 a

Extremum.

C: The graph of f has a low point and a high point.

D: For all x in the interval [- 2; 0] applies f (x) > 0.

E: The function f" meets the X-axis at the point x = -1

[slender lily]

If possible can you explain how to do in future problems

[fluid snow]

A: Monotoniesatz. Das besagt, dass die Funktion streng monoton fallend ist, wenn die Ableitung negativ ist

B: Was musst du denn prüfen, um herauszufinden, ob irgendwo ein Maximum oder Minimum vorliegt? ||Notwendige Bedingung: f'(x) = 0 + Hinreichende Bedingung: f''(x) > 0 bzw. f''(x) < 0 oder Vorzeichenwechsel von f'||

C: analog zu B

D: Welche Information geht beim Ableiten "verloren"? Denk an f(x) = x + 10000000 -> f'(x) = 1

E: Was bedeutet die Ableitung? f'' ist ja nur die Ableitung von f', was gegeben ist

[slender lily]

Wie darf ich verstehen, dass die Ableitung Negativ oder Positiv ist, also von der X-Achse gesehen negativ oder wie? (Bezüglich A)

[fluid snow]

Der Graph, der gegeben ist, ist schon die Ableitung, also schaust du einfach, ob es da unterhalb der x-Achse liegt (wenn ja, dann ist es negativ)

[slender lily]

Würde also f(x) so aussehen?

[fluid snow]

Ja, wobei es jede Verschiebung in y-Richtung haben könnte, die Information geht beim Ableiten ja veloren

Z.B. könntest du mit x^2 + 100 anfangen, die Ableitung ist dann 2x und wenn du dann versuchst, nur anhand 2x wieder auf die Originalfunktion zu kommen, hast du x^2, aber die + 100 ist verlorengegangen

Daher sagt man + C beim integrieren

[fluid snow]

Aber für diese Aufgabe musst du den Graphen nicht zeichnen

[slender lily]

Ich weiß aber ich habe keine Ahnung wie man aus der Ableitung der Ursprungs Funktion Informationen entnimmt, dass ist mir jetzt deutlich klarer geworden. Die Lösungen haben wir nämlich auch nichts gebracht.

Und noch eine Frage zu E warum hat f“(x) eine nullstelle bei -1?

[fluid snow]

Wenn du die Tangente an f' bei x = -1 anlegst, dann ist die waagerecht, also ist die Ableitung von f' (also f'') dort 0

[fluid snow]

Die Funktion aus der Ableitungsfunktion zu zeichnen macht man normalerweise bei der Integralrechnung, hier reicht es, wenn du anwendest, was du darüber weißt, wie man ein Maximum oder Minimum einer Funktion findet

Du kannst ja mal versuchen, die Sachen, die ich am Anfang geschrieben habe, durchzugehen und wenn etwas unklar ist, kannst du ja fragen

[slender lily]

.close

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[fading herald]

What’s happening with my inequality? I’m messing it up

[fluid snow]

How did x + 3 > 0 turn into x < -3

[fading herald]

well thats supposed to be the answer

[fluid snow]

The exercise is to determine the domain, right?

[fading herald]

yeah

determine domain

what i did is x > -3

but thats wrong

[fluid snow]

,w domain (x-4)/(sqrt(x^2 + x - 6))

Ok, the solution is right, let's see where the mistake is

[fluid snow]

It's not necessary that both are positive

[shadow knoll]

do you know wavy curve method

[fluid snow]

Both could be negative

And it'd turn positive

[fading herald]

nope

and never covered it

[fluid snow]

I prefer to imagine this as a parabola

(x+3)(x-2) has its zeros at x = -3 and x = 2

And it's opened upwards

So it will be positive on (-infty, -3) and on (2, infty)

[fading herald]

ok let me think

[fading herald]

tbh i dont really get what this is saying

[fluid snow]

This way is much faster than the algebraic one

[fading herald]

i have to show my work for my midterm though ;-;

[shadow knoll]

if you wanna go with the agebraic method just take the union set of the two possibilities

[fading herald]

i see thanks guys

you can close the channel now

i forget how to

[shadow knoll]

.close

[fading herald]

.close

[obtuse pebble]

Channel closed

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[fluid snow]

Depends on what's meant by that

Am I not showing my working here?

[fading herald]

ok gotcha

[fluid snow]

Unless your teacher told you to do it algebraically

(in which case you could tell him this way is faster when you have a quadratic and he should allow you to use this..)

[obtuse pebble]

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[stray timber]

is my work correct?

[tender stratus]

your final equation doesnt pass through the origin

[stray timber]

why?

i put (0,0) into the derivitive

[hexed needle]

put (0,0) in the final equation, it's not resulting a 0

[stray timber]

am sorry but am kind of confused wht u mean>

<@&286206848099549185>

.close

[obtuse pebble]

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[obtuse pebble]

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[paper swan]

On the sides AC and BC of a triangle ABC, isosceles on the base AB, consider two points P and Q respectively, such that the angle CPQ is equal to the angle ABC. just need help with this really quickly

[teal ledge]

What do you have to get for the result?

[undone yoke]

@paper swan

[paper swan]

that it is parallel

[undone yoke]

What is parallel ?

[paper swan]

oh mb it didn't send the whole question

[teal ledge]

you mean AB and PQ is parallel?

[paper swan]

yes yes

[teal ledge]

should they have to?

[paper swan]

yes they shold be parallel

[teal ledge]

please show me the full question

[undone yoke]

We have to prove that.

It seems

[paper swan]

yeah prove that it is parallel

drawing

like this i guess

[undone yoke]

Good. Observe that angle C is common for both.

Then angle CQP and angle CAB are also equal.

[teal ledge]

Do you have any geometrical conditions to show that two lines are parallel?

[undone yoke]

Since the angles subtended to a given line by two different lines are same, the two lines are parallel.

[paper swan]

oh is that a theorem

[undone yoke]

It's not, actually. It's a simple observation.

If the other set of angles was not equal, we couldn't have said anything.

[paper swan]

mhh

[undone yoke]

They basically wouldnt have been parallel.

[paper swan]

would you have been able to do it without the drawing or nah

[undone yoke]

I would have been. The question gave me enough data to be able to draw it myself.

[paper swan]

yeah but what if you couldn't draw it

my prof tells us that the drawing is there just to understand and you can't prove it just by observing the drawing but gotta write the proof

[undone yoke]

Tell your professor, in geometry, a drawing always comes first, even without the provision of data. Then comes the data itself.

[paper swan]

okok

could you help me out with this one

[teal ledge]

I think angle CQP is same as angle CBA is enough for parallel-ness

[undone yoke]

No. What is CPQ was not equal to CAB ?

[teal ledge]

they are corresponding angles

[undone yoke]

Then the lines would be intersecting. See @paper swan this is why figures are important.

[iron edge]

[paper swan]

yeahh you're right

so how anout this one

In a triangle ABC, let BP be the bisector of the angle B. From P draw the parallel to BC; call Q the point in which meet the side AB. Always from P draw the parallel to AB until it meets the BC side. Prove that the triangle PBQ is congruent to the triangle PBR

[undone yoke]

Okay don't give a figure now.

[paper swan]

alright

[teal ledge]

what is R

PR with BC?

[undone yoke]

Yeah exactly.

How is it made ?

Perpendicular subtending. Random intersection?

How is it?

[paper swan]

r is the point that meets CB

[teal ledge]

then PR is parallel to AB?

[paper swan]

yes

[undone yoke]

,rcw

[undone yoke]

@paper swan

[teal ledge]

with BP in common

[paper swan]

oh they're congruent because all their angles are?

[teal ledge]

should include that BP is in common, so ASA congruence

[teal ledge]

I still think CQP is same as CBA, as CQP=CBA=CAB

[undone yoke]

Haa that'll be there due to the isosceles property. But we were discussing what if it wasn't isosceles ?

[paper swan]

yeah in the first question it is given it is isosceles

[paper swan]

could you explain to me this better rq

[teal ledge]

because this doesnt ensure that they are parallel if no isosceles

I think the focus of what I was talking is different from what you understanded as. My fault, I am sorry

[teal ledge]

Have you learned any specific conditions to show that two lines are parallel?

[paper swan]

If two lines cut by a transversal form with it:
two alternate (internal or external) equal angles, or
two equal corresponding angles, or
two additional conjugate angles (internal or external),

the two linesa re parallel

i learned this

[teal ledge]

I am not 100% sure on what solomaniac wanted to say, but I think he wanted to show that CPA=CAB and CQP=CBA

What I wanted to say is as CQP=CBA, two lines are parallel because they are two corresponding angles, and they have same angle

[paper swan]

yeah the two lines are parallel because when cut with a transversal they form equal corrisponding angles

[paper swan]

yeah you're right

[teal ledge]

I think me and solomaniac had different axiom to show parallel-ness

no big problem I guess

Sorry for confusion again.

@paper swan Do you have any other questions?

[paper swan]

yes

[paper swan]

so they have bp in common

and what else to determine they're congruent?

[teal ledge]

Can you explain ASA congruence?

Or do you know about ASA congrunence?

[paper swan]

yeah when two triangles have a side in common and the two angles on that side are congruent

[paper swan]

so how do we prove that

[teal ledge]

To add more QBP is same as RBP because BP is angle bisector

So QBP=RBP, BP in common and QPB=RPB

[paper swan]

i don't understand how you can prove that QBP = RBP

[teal ledge]

because BP bisects angle QBR

which means QPB and RPB should both be half of QBR

[paper swan]

so because they form alternate internal angles?

or that doesn't have anything to do with it

[teal ledge]

oh sorry typo issue

QBP and RBP both should be half of QBR

[paper swan]

ah okay okay

[teal ledge]

Actually you dont have to say QBP=RBP for congruence

not necessary

QBP=RPB(alternal angle), QPB=RBP(alternal angle) and BP in common

so two angles, and side between those two angle is all congruent

which means ASA congruence for triangle

[paper swan]

so because they are alternate angles?

[teal ledge]

yes

[teal ledge]

this with BP in common is enough for congruence

[paper swan]

yeah but it says that bp is the bisector of just the angle b

not of the angle p

[teal ledge]

But as question is asking for PBQ and PBR, using bisection is appropriate now I think

[paper swan]

what do you mean?

[teal ledge]

angle B is angle QBR

[paper swan]

yes

bp is the bisector of angle QBR

and it divides it in half forming two congruete angles

but what about angle QPR

how do you know that it is also the bisector of that angle

[teal ledge]

As I know when writing dowm congruence, the order of triangle name is important

[teal ledge]

because QPB=PBR=QBP=BPR
alternal,bisection,alternal for equalness

[paper swan]

yeah but how do you know that QPB=PBR=QBP=BPR

[teal ledge]

QPB=PBR for same alternate angle
PBR=QBP for bisecting QBR with BP
QBP=BPR for same alternate angle

[paper swan]

ohh okay yeah now i get it

[paper swan]

can you help me with one last thing?

[teal ledge]

okay

[paper swan]

Given two parallel lines r and t, and a transversal t that meets the line r in P and the line s in Q, draw a line through the midpoint M of PQ that meets r in R and s in S. Prove that PR = SQ

[teal ledge]

Also ASA congruence

MP=MQ

RPM=SQP

RMP=SMQ

so two triangles are congruent

whicb means PR and QS should be same in length^

[paper swan]

no not the triangles

the lines PR and SQ

[teal ledge]

yes

[paper swan]

OHH

right

[paper swan]

how do you prove that tho

oh because it's the midpoint

[teal ledge]

yes

[paper swan]

what about this

[teal ledge]

alternal angle

[paper swan]

ohh right

[teal ledge]

they have same angle because r and s is parallel

[paper swan]

you mean r and t?

or R and S

[teal ledge]

"Given two parallel lines r and s,..." in your question

[paper swan]

ohh mb it was supposed to be a t

i thought of something like this

the angles QMS and RMP are congruent because they're opposite

[teal ledge]

you are right. Only angles are not enough though. IMO It's good to look for at least one side congruence

[paper swan]

ohh okay

so if all three angles are congruent it isn't enough

[teal ledge]

if all three angles are same, but dont know about sides, then you say those two triangles are similar

but not enough for congruent

[paper swan]

ohh i see

thank you man

[teal ledge]

np

[paper swan]

you can close the channel now i'm good

[teal ledge]

you should

!done

[obtuse pebble]

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

[paper swan]

.close

[obtuse pebble]

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[dark idol]

BDF r wrong

[dark idol]

Help

[undone yoke]

Yeah they are. You haven't understood what the question was asking.

It was asking $5^+, 6^-$

[warm shale]

Solomaniac

[dark idol]

Ik that

[undone yoke]

As for f, both the limits of $6^+, 6^-$ are going to -infinity.

[warm shale]

Solomaniac

[dark idol]

How

It looks like 6- doesn’t even go to 1

Same for 5+

[undone yoke]

6- is going to -infinity.

Same for 5+

[dark idol]

.close

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[random olive]

"(The old man in the box) A cubic box has a corner at the origin and edges OA, OB
and OC with others in the points A = (2, −2, 1), B = (1, 2, 2) and C =
(−2, −1, 2). A dot-shaped old man is at the point with coordinates
(0.9, −0.9, 4.5). Is the old man in the drawer?
Tip: Transfer to a suitable new base. Rephrase the question 'lie in the drawer' in ¨
algebraic terms. What finesse does the change to a new ON base offer?"

[random olive]

does anyone know how to solve this?

[obtuse pebble]

@random olive Has your question been resolved?

[random olive]

.close

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[dapper wing]

why is this wrong?

[dapper wing]

pi/3 - 1 = .047

how do i convert that to radians

or is it (pi-3) / 3

[obtuse pebble]

@dapper wing Has your question been resolved?

[dapper wing]

ty

[obtuse pebble]

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[stray timber]

i am confused about this question i found the derivative

do i just equal the derivative to 0

[brave bramble]

You'll need to find dy/dx before anything else

But yeah, dy/dx = 0 at those points

[stray timber]

but i does not give me any points

[brave bramble]

Correct, you've got to find them

[stray timber]

ah ok

thx

[obtuse pebble]

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[heavy chasm]

how to solve this

[brazen gorge]

how much did he drive from 8:00 am to 1:00 pm?

how many kilometers

[heavy chasm]

uhh

[brazen gorge]

first he was 510 km away

now he is 105 km away

[heavy chasm]

he drove

405 km

[brazen gorge]

yep

how many hours is from 8 am to 1 pm?

[heavy chasm]

5

[brazen gorge]

yep

so in 5 hours, he drove 405 km

[heavy chasm]

yes

[brazen gorge]

do you know the speed formula?

[heavy chasm]

distance over time?

[brazen gorge]

yes

[heavy chasm]

ok

[brazen gorge]

you have distance, you have time. result will be your answer

[heavy chasm]

81

[brazen gorge]

yep

that's it

[heavy chasm]

can u help with another one

[brazen gorge]

yes

[brazen gorge]

okay

how many marbles in total are there?

[heavy chasm]

24

[brazen gorge]

oh it says 24

lol

[heavy chasm]

yea

[brazen gorge]

what's the formula for probability?

or like calculate the probability of drawing a red marble

[heavy chasm]

uhh

[brazen gorge]

without adding new ones or anything

[heavy chasm]

8/24

[brazen gorge]

yes

let n be the number you need to add

so you get $\frac{8+n}{24+n}=\frac{3}{5}$

[warm shale]

artemetra

[brazen gorge]

solve for n from here

hint: cross-multiply

[heavy chasm]

multiply 8+n times 5?

[brazen gorge]

uh

that will be the lhs

so yeah

[heavy chasm]

what's lhs

[brazen gorge]

left hand side

[heavy chasm]

72n/40n?

[brazen gorge]

huh?

[heavy chasm]

or does the n like

stay away

[brazen gorge]

you should do $5(8+n)=3(24+n)$

[warm shale]

artemetra

[brazen gorge]

that's what i meant by cross-multiply

[heavy chasm]

ight lemme solve

wait how does 5 end up on the left side

oh nvm

u got rid of both denominators

[brazen gorge]

yes

[heavy chasm]

40+n=72+n

oh wait

[brazen gorge]

no

[heavy chasm]

ignore that

5n+40=3n+72

[brazen gorge]

yes

do you know how to proceed from there?

[heavy chasm]

lemme think

subtract 40 to put on right side then divide 3m

3n*

5n/3n=32

i thibk i messed up

wait

5n=96n

n=96n/5

[brazen gorge]

no

[heavy chasm]

dammit

[brazen gorge]

subtract 3n from both sides

[heavy chasm]

ohhh

u right

n=16

that was easier

.close

[obtuse pebble]

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[tawdry talon]

If we let $\xi$ be random variable which denotes amount of points on the dice, then $E\xi^2=1^2P(1)+2^2P(4)$?

[warm shale]

kitten

[timid silo]

is that the q? because that doesnt seem correct. what about the other options:2,3,5,6?

[tawdry talon]

Isn't P(3^2) = 0 and on because we have 6-sides dice?

[timid silo]

No. We aren't squaring \zeta inside the probability.

$$E[\zeta^2] = 1^2 P(1) + 2^2 P(2) + 3^2 P(3) + 4^2 P(4)+5^2 P(5) + 6^2 P(6)$$

[warm shale]

swimmingland

[tawdry talon]

Ohhh, I see, thank you 😊

.close

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[timid silo]

Sure thing! 🙂

[obtuse pebble]

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[dark idol]

Help

[dark idol]

U can move around any of these boxes

[obtuse pebble]

@dark idol Has your question been resolved?

[dark idol]

<@&286206848099549185>

<@&286206848099549185>

[wide sequoia]

Hey

[soft grove]

What

[wide sequoia]

how can i help

[dark idol]

Idk how to do this

[wide sequoia]

please dont spam us

[soft grove]

What is 2+1

Helpers pls help

[wide sequoia]

IDK

[soft grove]

NOOOO

[wide sequoia]

BrO ThAT tO HArD

[soft grove]

It's the hardest question in the world

[wide sequoia]

WAIR

ITS 3

IT MUST BE 3

[soft grove]

YEAH UR RIGHT

[wide sequoia]

OMG

IM ALBERY ENSITE

[soft grove]

WE ARE GENUISES

[wide sequoia]

IK

[dark idol]

Ok so u guys don’t know it either

[wide sequoia]

Let me check

Its ask the teacher for extra credit

lol

@dark idol

What grade are you

[dark idol]

Idk why that’s there

[wide sequoia]

Joke anwser

[dark idol]

Ok so how I do it

[wide sequoia]

What grade are you, and what math class is this

[dark idol]

It’s calculus college

[wide sequoia]

Okay

Yeah

This is above my level

[dark idol]

Bruh

[wide sequoia]

Im only grade 11

[dark idol]

No problem

.close

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[sand forum]

I don't understand the solution. Why are they using inclusion exclusion for 4-sets, but getting so few terms? There should be 15 terms to work out

[drifting wraith]

they have 14 terms

4, then 6, then 4

they are missing the last one

[sand forum]

so, there is a problem with the solution?

I see they have grouped them now, and that there is 4 + 4 choose 2 + 4 choose 3 = 14 terms ...

[drifting wraith]

i'm not sure, i would say yes

[sand forum]

is there another approach than inclusion exclusion to this problem

[drifting wraith]

oh i see

the last term is 0

7 cards can't fit 4 pairs

[sand forum]

oh yea, there is only 7 cards

I missed that some how

[drifting wraith]

sure, you can just add 3 terms

[sand forum]

?

[drifting wraith]

hm

no nvm

[sand forum]

probability of an ace and king of the same suit in a 7 card draw from 52 standard card deck

can we just compute a conditional probability maybe

hmm no maybe that is harder

maybe I have a skill issue since I didnt try to group the terms like the solution did which made the work more expensive for me

[obtuse pebble]

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[long hedge]

$81\int :sec^5\theta -sec^3\theta :d\theta$

[warm shale]

putridplanet

[long hedge]

i know this is a dumb question but you can't just subtract the sec^3 from sec^5 like that to get sec^2 because you are taking the integral of both

that would only work if there was no integral

right ?

ok but that would just make the answer tan +c

and my teacher didn't do it that way she took the individual integral of both

oh i thought you said you could

ok

.close

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[faint cape]

1. Question: Complete the table so that sine theta is expressed in terms of the functions given across the top. (Cot theta)

[faint cape]

my work so far

I am so confused on what this even wants me to do

I know to use trig identities but like idek what to do other that 1/cos

these are my options

more of what i tried

im actually working on the sec one first my bad

<@&286206848099549185>

do i ping at 15 min intervals?

<@&286206848099549185>

[obtuse pebble]

@faint cape Has your question been resolved?

[obtuse pebble]

@faint cape Has your question been resolved?

[obtuse pebble]

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[tawdry saddle]

im wondering if i can cancel out the (x + 3) in the numerator and denominator. Usually it would be obvious but there is a coefficiant of 2 in the denominator and i dont know if that matters or not. Can anyone let me know if its still ok to cancel out the (x + 3)?

[latent walrus]

depends what youre doing

but the 2 isnt an issue to whether you can cancel

[tawdry saddle]

ok

Im simplifying a fraction

[latent walrus]

i suppose you can then

[tawdry saddle]

This is the full question

[timid silo]

you can cancel them

[tawdry saddle]

ok

[latent walrus]

if youre doing graph stuff, just keep in mind that if x is -3 you have a hole and its undefined there

[latent walrus]

since its a removable discontinuity

[tawdry saddle]

yeah Im very confused on the hole thing

[latent walrus]

a hole happens when the function is undefined at a point but its limit has a value there

[tawdry saddle]

if it cancels out its a hole?

[latent walrus]

in a limit the x-3 will cancel

so the limit is defined at -3, but the function isnt

hence a hole

thats why i meant be careful in cancelling

context is important

[tawdry saddle]

now when i find the verticle asymptotes

do i keep the x + 3

or would it only be 1 asymptote then

Right now this is all I have

[latent walrus]

would just be the x-3
we know that (x+3) gives us a hole rather than an asymptote, so just the x-2

you want the limit to be undefined

the x+3 disappears in limits

[tawdry saddle]

so what do i put for holes

[latent walrus]

function undefined and limit defined

[tawdry saddle]

Do I just write the x + 3?

[latent walrus]

see where the function is undefined, check if the limit is or not

[tawdry saddle]

I remember my teacher showed us some equation but I don’t remember it

[latent walrus]

if it is then its a hole, if its not then its an asymptote

[tawdry saddle]

so if im cancelling out x + 3

is the hole x - 3 or x + 3

i mean x = 3

or x = -3

thats how my teacher wants us to write it

[latent walrus]

see where the function is undefined:+3 and -3
look at the limits, x+3 cancels out, and so the limit at -3 is defined, this must then be a hole
look at +3, the limit is also undefined since x-3 doesnt cancel, this is an asymptote

hole: f undefined, lim defined
asymp: f undefined, lim undefined

[tawdry saddle]

when i put the equation in my calculator both -3 and 3 are undefined

does that mean they are both holes?

[latent walrus]

... no, what are you talking about

have you read what ive actually said?

[tawdry saddle]

Im trying but i have no idea what it means

my teacher didnt teach us like this

[latent walrus]

this is as simple as i can put it really

yes f is undefined at +3 and -3, but the limits arent necessarily

[tawdry saddle]

ok

my teacher never told us what a limit is

or at least she must havbe explained it differently

So i dont reallyt know what you mean when you are talking about limits

[latent walrus]

oh, you dont know limits

okay, if the part that makes the function undefined at a point can cancel out, then that is a hole, if it cant then its an asymptote

[tawdry saddle]

okay so x+3 is a hole then right?

because i canceled it out

[latent walrus]

-3 is a hole yes

[tawdry saddle]

yes x = -3

thats what i wanted to know

if (X + 3) chances to x = -3

changes*

[latent walrus]

you lost me there

[tawdry saddle]

i canceled out the (x + 3)

[latent walrus]

x+3 doesnt change to x=-3

[tawdry saddle]

but i canceled the x + 3 and you said x -3 is a hole

[latent walrus]

you need to stop writing things like this, its confusing

x=-3 is a hole yes because solving (x+3)=0 gives x=-3 and the (x+3) cancels in the function

you lost me at the word 'changes' because it makes no sense

[tawdry saddle]

ok sorry

[latent walrus]

if you mean x=-3 then write that, if youre referring to the x-3 in the function then write x-3

there has to be distinction

[tawdry saddle]

this is how my teacher taught us to solve holes

[latent walrus]

um, what was the original function

[latent walrus]

okay, that makes sense, i very much dont like how its written though

that last part is not f(x)

its a different function

but yeah, theres a hole at 0, because the function is undefined but the part making it so can cancel

[tawdry saddle]

here is my work and the original function

sorry its on its side

[latent walrus]

hold on, actually this is just nonsense

it would be fine if using limits but this isnt being done

the point (0,2) is not even part of f

sure theres a hole at 0 but saying f=-2 there?? Why are they doing that

[tawdry saddle]

Dude I have no idea this is for a grade though honestly I just want to get a good grade

I barely even understand the mechanics of this

[latent walrus]

the limit at 0 is 2, not f itself

f is undefined at 0

teacher cutting corners

[tawdry saddle]

ok

i dont really understand

[latent walrus]

please dont write that f(-3) is that

its fine to say the hole is at that point you found, but dont say that its the value of f there

[tawdry saddle]

so what do i put

[latent walrus]

its a notation issue, it doesnt make sense without using limits
you can say the hole is located at (-3,-16/(-12)) ie (-3,4/3)

just dont say f(-3)=4/3
if you want to be classy, say
$$\lim_{x\to{-3}} f(x) =\lim_{x\to{-3}} \frac{3x-4}{2(x-3)}= \frac{4}{3}$$

[warm shale]

AℤØ

[latent walrus]

that would be the most correct

[tawdry saddle]

Is there supposed to be 2 separate or did I graph it wrong

[latent walrus]

though i get that may be beyond your course

you cant see a hole on a graph

[tawdry saddle]

Yeah I’m a sophomore just trying to get a good grade on this

[tawdry saddle]

Ok but is the 2 separate lines ok

[latent walrus]

yeah of course

where there is a hole we often draw a hollow circle at that point

[tawdry saddle]

Ok

I’m confused on how to graph this

Like on paper

I need to use that x and y chart

And idk which values to pick

[latent walrus]

essentially just this, but put the hollow circle on -3

0 is essential, the rest you can just pick at random

but id label your axes first

might help in choosing

[tawdry saddle]

3 is also an error

What is that about

[latent walrus]

3 is the asymptote

[tawdry saddle]

Ok does that matter when I graph

[latent walrus]

yeah, you have to draw it

check which way its going from each side

[latent walrus]

you can see on the right it goes up, check some value like 3.1 or something on the left maybe check 2.9

[tawdry saddle]

Are these values ok

[latent walrus]

instead of error, write undefined

put those can work

you dont have space on your graph to get to 5 though

[tawdry saddle]

What do you mean

I have plenty of space

[latent walrus]

you really want some closer points the the undefined bits to see the behavior there

[latent walrus]

okay okay

[latent walrus]

you havent labeled your axes with any numbers

no one can tell the scale

but its definitely wrong anyway, your asymptote isnt vertical

thats important, itll help to actually mark where x=3 accurately is

[tawdry saddle]

Well idk how to fix it I checked every step

It’s getting late where I am

Honestly I’m trying to finish this

[latent walrus]

you need to label your axes, i told you

you cant make an accurate graph if you dont know where your values are

[tawdry saddle]

Oh the bottom one can’t pass the 3

Because that’s the asymptote

[latent walrus]

the function cant cross the line x=3 at any point

[tawdry saddle]

Got it

I fixed and laboed

Ok next question

[latent walrus]

[tawdry saddle]

i was very confused when the teacher explained this

[latent walrus]

i presume youve seen the rational root theorem if this came up

[tawdry saddle]

probably

i honestly had no idea what was going on during this

i know the rest pretty well but this possible rational roots thing

i dont get it

[latent walrus]

this is an overview and example

actually ignore that its wrong oops

this one is fine

[tawdry saddle]

so you divide the factors of the first coefficient with the factors of the last one?

and thats all the possible ones

[latent walrus]

factors of the constant by the factors of the coefficient of highest order x yeah

part b will probably involve some polynomial division of some form

[tawdry saddle]

so what do i do with all those possible ones

i think i remember my teacher talking about substituting but i may be wrong

[latent walrus]

you sub them all into the polynomial to see which ones are actually roots

[tawdry saddle]

how do i know if its a root

if it equals 0?

[latent walrus]

yeah

[tawdry saddle]

damn i gotta substitute all of those

[latent walrus]

heh

[tawdry saddle]

do i need to do the +- seperately

[latent walrus]

yup

[latent walrus]

wow just realised this gave you all your roots lol

oh well, at least its relevant

[tawdry saddle]

for my roots

i got 1, -1/2, 1/3, 3

well at least its 4 roots

no idea if its right or not

next question

😭

[latent walrus]

looks worse than it is, just factorise everything you possibly can

[tawdry saddle]

first i need to swap the numerator and denomerator of the 2nd part right

and change it to multiply

[latent walrus]

sure

[tawdry saddle]

problem is so long even photomath couldnt recognize it

i know how to do this its just going to take me so long and i want to go sleep

im just going to skip for now

@latent walrus isnt this just horizontal asymptote?

or is there something more

Here’s my work

[obtuse pebble]

@tawdry saddle Has your question been resolved?

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[gritty grotto]

i dont where to go after i put it into the matrix format

[gritty grotto]

ik im supposed to get 3 0s and lke check but like what is the question asking

im just confused in general

[mystic dagger]

you must determine range of k

[gritty grotto]

so will it be like k>n?

[mystic dagger]

i think maybe

you can get a, b, c by k using taken assumptions

[gritty grotto]

where do i go from here

[mystic dagger]

solve them using [-4, 2, 2k]

do you solve

[gritty grotto]

wdym

oh

using the answers

how do i solve

i dont get

how do get rid of the k^2

[mystic dagger]

A[a; b; c] = [-4, 2, 2k]

[a;b;c] = A^(-1)*[-4, 2, 2k]

[gritty grotto]

wait hold up

how d u get abc

and what are you using

im not too sure

[mystic dagger]

using formula

[gritty grotto]

which?

[mystic dagger]

Matrix

[gritty grotto]

?? trheres formual

[mystic dagger]

are you university student?

[gritty grotto]

no

i am

in year 12

15 years old

what does it have to do?

with that

[mystic dagger]

oh, it is formula which can learn in university

[gritty grotto]

how do i relate that tgo this?

[mystic dagger]

i will think some other ways

[gritty grotto]

ah ic ic? i dont think i will be able to remember it

[gritty grotto]

thanks

[mystic dagger]

oh, i think it is not difficult problem.

this problem is which type in your textbook

[gritty grotto]

i thnk we use gausean

method

gaussian

elimination

[mystic dagger]

oh

so this problem means that get appropriate k to not exist proper a, b, c

right?

[gritty grotto]

i think

im lost

i learnt system of eqn today

[mystic dagger]

i think determinant value of matrix should be 0

[gritty grotto]

what is determinant

idek at this point

[mystic dagger]

oh, no

this is also unversity's

[gritty grotto]

ok... so this is

the system

of eqn

what should i do with it?

substitute kc?

solve for kc?

[mystic dagger]

substitute

[gritty grotto]

ok

but then what?

use discriminant?

[mystic dagger]

and also substitue ak^2

[gritty grotto]

so should i solve for ak^2

and then sub it into q1 or 2

[mystic dagger]

use eq1, eq2 bk = 3

right?

[gritty grotto]

yup

hi?

<@&286206848099549185>

[fierce elbow]

u gotta sovle these 3 equations?

[mystic dagger]

k must be -2 or 2

what is correct answer?

@gritty grotto

[gritty grotto]

yes

thats correct

but how do u do it

@mystic dagger

[mystic dagger]

-2 and 2 is all ok?

[gritty grotto]

yup thats right

but how do i get the answer

without using uni stuff

[mystic dagger]

what is uni stuff?

i think there might be more solutions, i guess 4 solutions, but it is too difficult to calculate them

[gritty grotto]

is there a way to do it just through pure substitution

basics?

what did u use?

[mystic dagger]

you only use gaussian elimination

and finally you can get:

k^3-4k | 2k^3- 6k+4

so solution will be +2 or -2

got it?

hey @gritty grotto

[gritty grotto]

do u have working i can follow?

sorry im not that good at visualising it

[gritty grotto]

wats the thing in the middle

|

[mystic dagger]

gaussian elimination

wait a minute

i will send answer

[gritty grotto]

okk

i learnt this method today

im not even sure myself

[mystic dagger]

first is your image

ok?

[gritty grotto]

yuh

[mystic dagger]

second is

ok?

simply substitute eq1 from eq2

hi?

[gritty grotto]

wait

im tryna understand

oh is

oh ic

[mystic dagger]

third you multiply 4/k^2 to eq1 and substitude from eq3

let 4 goes 0

ok?

[gritty grotto]

wait lemme try

wdym by substitute from eq 3

so eq3-eq1?

[mystic dagger]

eq1 remains unchanged

[gritty grotto]

yeah i just write it out

[mystic dagger]

you must do eq3 - eq1*(4/k^2)

[gritty grotto]

yup

ok

like th

[mystic dagger]

yeah

and like this eq2. and eq.3

[gritty grotto]

wdym?

[mystic dagger]

eq3's second term goes 0

ok?

[gritty grotto]

eq2-eq3?

[mystic dagger]

... ... ...
0 ... ...
0 0 ...

[gritty grotto]

eq3-eq2?

[gritty grotto]

how do i get that im sorry so sorry

[mystic dagger]

eq3 - eq2*(2+4/k)/2k

[gritty grotto]

wait