help-0

[tidal fog]

its 3 am

im so done with it

like

for a i got 2 3 0

but idk how im suposd to get the second part

[vale wigeon]

deadline?

[vale wigeon]

show your work?

[tidal fog]

this isnt an assignment

its hw

for my test tmrw at 4pm

i cant take a pic im in the dark my mom will scream at me if i turn it on

[vale wigeon]

... well like

did you get an all zeros row at least

[tidal fog]

no wym

i got]

1 0 0 | 2
0 1 0 | 3
0 0 1 | 0

by doing rref

which is so anoying

and takes me slong

is there any tips for that

[surreal meadow]

we can’t help if we don’t know what you did wrong. send your work if you have some

[vale wigeon]

they already said they are unable to

[surreal meadow]

oh

[lone heart]

@tidal fog Has your question been resolved?

[lone heart]

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[rotund shoal]

I am a bit confused with the second part of the working out

From "For 2 distinct roots" onwards

I believe there is an error in factorising in the 2nd line

However the last line corrected it.

My main concern is how to determine that the arrows point out

[granite badger]

hello again

[rotund shoal]

In my original answer, I got m > -2, 6

[subtle birch]

discriminant must be greater than 0

[rotund shoal]

I understand that.

[rotund shoal]

^

Do I need to recognise that it is a graph of a parabola

because by just solving the inequality, I get m > -2,6

[subtle birch]

NO

graph it

[granite badger]

i guess you just have to remember this pattern for 1 sol, 2 sol or no sol

[rotund shoal]

I am familiar with the solutions and recognised that there are roots at -2 and 6

However getting from here to here

[granite badger]

they made a mistake

[rotund shoal]

i believ it should have been (m+2)

[granite badger]

(m-6)(m+2)

[rotund shoal]

but even at that, I don't understand how the m+2>0 becomes m < -2

[subtle birch]

The answer is correct but steps are wrong

[subtle birch]

do you know how to use wavy curve method?

[rotund shoal]

As stated in the original message, I have determined this.

[rotund shoal]

No

I may know the method, just not the name of it

never heard of that

[granite badger]

i’m not too sure myself but the way i do it is just the smaller number is m < number and the bigger number is m > number

[rotund shoal]

I see, but thats with the assumption that you have a parabola

I was just wondering if that was absolutely necessary to recognise before stating the domain

Since originally, I waved over the fact that it was a parabola and just did m>6 since that outrules m>-2

I was merely looking a the inequality that I made

[granite badger]

im not sure myself in this too much as i have never bothered to venture deep into solving problems like this

i usually find roots and apply what i need to do to achieve x sols

[rotund shoal]

Gotcha, I suppose next time i just have to recognise early that I am solving for a parabola

Anyway, thanks for clearing it up

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[jade cobalt]

I'm so confused why my teacher divided the leading coefficients

[jade cobalt]

Ok so I graphed the first equation as seen in red. The limit is clearly -infinify

Then I changed the power to 6

When I cancel them out, it becomes -x^3

If I were to plug infinity into this, it would just become -infinity again

Wait Holy shit I'm stupid

Nvm

As x approaches infinity

It goes to negative infinity

But not as x approaches -infinity for the power of 6

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[jade cobalt]

.reopen

[lone heart]

✅

[jade cobalt]

Wait but hold on

If I cancel the leading coefficients out, I get x^3

When I plug infinity into that, I should get infinity, no?

So why do I get -infinity on both sides

Is it because when I factor it I get $\frac{x^2(4x^2 + 5)}{x - 4}$

And then the 4x^2/x = 4x

But even then

Both are positive

I'm so confused

[ocean seal]

TreetopperJoey

[jade cobalt]

Why is the limit as x approaches infinity in $\frac{4x^4 + 5x^2}{x - 4} -\infty$ ?

[ocean seal]

TreetopperJoey

[jade cobalt]

Nvm it isn't 💀💀

[hushed locust]

try using l'hopital's rule

[jade cobalt]

.close

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[jade cobalt]

Oh I'm not in calculus yet

Sorry

Precalc

My friend told me about that

He said it's super easy

But seeing as my test is in 2 hours probably not worth learning

[hushed locust]

yeah it just looks like that up close

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[limpid stone]

how are these two both standard deivation

it doesnt make sense

[vale wigeon]

they're stdevs of different things

trying to think of a way to put the latter into words but i cannot find the right words

[lone heart]

@limpid stone Has your question been resolved?

[limpid stone]

how would i tell which to use in waht situation

[vale wigeon]

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[light oyster]

Any idea on part(b)?

[astral briar]

the way the circle is written is pretty unusual

see if you can write in the form

(x-a)^2+(y-b)^2=c

[light oyster]

Don't get what it means to be OQ:QR = 1:4

[naive crystal]

The circle can be written as (x-3)²+(y-6)²=25

[astral briar]

quick one

[light oyster]

yes

[naive crystal]

Ration of there length

[light oyster]

I know but how can i calculate the m

m = 1/2 or -3/2, when i put y=mx into C and calulate using 5a^2 = 36b

[astral briar]

o is at the origin btw

[light oyster]

ye

oh btw use of graphing calculator in Hong kong exam is prohibited so actually i draw by hand when i'm doing this question in my university entrance ex am

[astral briar]

im not using a graph dw i just wanted to show ykwim

[light oyster]

ok

[astral briar]

https://tenor.com/view/travis-scott-headbang-shake-sicko-mode-smh-gif-17416858

[naive crystal]

Only if we could get the coordinate of Q

[astral briar]

if the line y = mx intersects the circle at a point p, that would mean that

(x-3)^2+(mx-6)^2=25

also

if you are familiar with the section formula (i think it can be used here)

[naive crystal]

Yep

So we got one coordinate of Q

(0, something)

Now I think that

We can use distance formula for the radius

Like

[astral briar]

why 0

we gotta show that brah

[naive crystal]

$\sqrt{(3-0)^2 + (6 - a)^2}=5$

[ocean seal]

Random Guy

[naive crystal]

Yep

Solve for a and we get the coordinate of Q

Now use slope formula to calculate the m

[astral briar]

why is Q on the y-axis

[naive crystal]

✔️

[astral briar]

you have not explained why

[naive crystal]

See

x1 and x2 are (0,0) agreed??

[astral briar]

https://media.discordapp.net/attachments/958029650460827699/1070947549453766686/38.gif

[naive crystal]

That's why they called it origin bruh

@light oyster

[astral briar]

how is that x1 and x2

[naive crystal]

Did you get it??

You yourself mentioned here

Bruh

[astral briar]

i think you are applying the formula incorrectly

[naive crystal]

Oh ye

I'm dumb

F

[astral briar]

this is why india's workforce is mainly in tertiary sector

[naive crystal]

Bruh

[light oyster]

Irrelavant to section formula, seems i've found the answer

[astral briar]

https://tenor.com/view/travis-scott-headbang-shake-sicko-mode-smh-gif-17416858

[light oyster]

$x^2 + x\frac{-12m-6}{m^2+1} + \frac{20}{m^2+1} = 0$

[ocean seal]

小馬

[light oyster]

damn still need to use section formula cuz O,Q,R are collinear

[naive crystal]

So basically The coordinate of Q is (x2/5,y2/5) now if we apply distance formula between Q and center we should be able to get the answer

[light oyster]

By Section Formula :
$$\frac{4(0)+1(x_2)}{4+1} = x_1}$$
Hence,$ x_2 = 5x_1$

[ocean seal]

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

[light oyster]

Then use 5a^2=36b, solved

[naive crystal]

What is the value of M then

[light oyster]

slope

[naive crystal]

Ik

But I was asking for the actual answer

[light oyster]

3/4

based on my calculation

[naive crystal]

Let me solve it using my method

Wait a min

,w (3-x/5)^2+(6-y/5)^2=25, (3-x)^2+(6-y)^2=25

[ocean seal]

Results provided by WolframAlpha

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

Requested by Random Guy#5548

[naive crystal]

Ummm....

Yep

I also got 3/4

@light oyster

[light oyster]

nice

[naive crystal]

My approach was a little bit high school friendly

But idk about your

[naive crystal]

This is disaster 💀

[light oyster]

not really, i used the results from (a)

[lone heart]

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[golden arch]

What is the differential equations that matches the given vector field. Choose the correct answer:

How would I start solving this?

[safe tartan]

Look for gradients of 0 on the vectors

[golden arch]

i just integrated it and got the correct answer

is this also a way to do it

[safe tartan]

Integrated each answer?

[golden arch]

well the vector field is circle shaped

so I integrated the first one and got a circle as an answer

[safe tartan]

I mean if it works

[golden arch]

.close

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[cerulean canopy]

Trigonometry (ASTC)
i really dont understand pls help 😭

[pulsar glen]

sin-1

[cerulean canopy]

yes

[pulsar glen]

then create 2 equations

[cerulean canopy]

sin-1 (3/4)?

[pulsar glen]

yes

then separate into 2 equations

[cerulean canopy]

howw

[pulsar glen]

because -3/4 cuts the sin curve in 2 different positions

[cerulean canopy]

so....

i dont get it

[pulsar glen]

when u plug in calc u get -48. smthn

then

between 0-360 you can see we have to

add 180 + 48.smthn

[cerulean canopy]

but they want me to use the ASTC thingy

[pulsar glen]

i use the calcultor to convert it to that

[cerulean canopy]

can somebody try solving and sending their ans

no im not tryna get you to do my homework ive got like 30 questions i just need 1 example to work backwards and ill figure out myself

(i always work backwards and figure out myself)

plss 🥺

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[craggy token]

Given two positive different rationals $x, y$ such that $$w=\frac{x+\sqrt{y}}{y+\sqrt{x}}$$ is rational. Prove that $x$ and $y$ are the square of rationals.

[ocean seal]

$nichoals$

[craggy token]

How do I solve this?

[lone heart]

@craggy token Has your question been resolved?

[craggy token]

.close

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[finite coyote]

I am just confused, here am I supposed to remove the denominators of both equations to find if a1 and a2 equals b1 and b2 but not c1 and c2?

[torn elk]

You can make b so firs equation LHS is equal to second equation LHS, so that would make impossible to solve cause they have both different RHS

to do that, you simply divide second equation coefficient with the first one, and the result multiply by the first coefficient of the first equation that multiplies x and with the second one.

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@finite coyote Has your question been resolved?

[wet nest]

a/a1=b/b1≠c/c1

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[young scaffold]

are those all the trigonometric and hyperbolic variations of tan ?

[worn fox]

depends what you mean by "variation"

[young scaffold]

all the other functions that involve tan or tanh

[worn fox]

pretty vague still

tan(x) + 5 "involves tan

(tan(x))^2 involves tan

[young scaffold]

More complex functions that are just formed from the basic trigonometric or hyperbolic functions

[worn fox]

"more complex"? is (tan(x))^2 not more complex?

sqrt(tan(x))?

e^(tan(x))?

like i dont think you're accurately saying what you want

[young scaffold]

.close

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[proud moth]

Surely I can’t disprove the statement

Like I can’t think of a single counter example

[plain flame]

Non zero integers

So you can let one be negative

[proud moth]

Wait how do I find a a gcd with just 1 negative value

Uhhh could you elaborate please I don’t understand

[plain flame]

Negative integers are divisible by positive integers

And the GCD, LCM are defined to be positive

[mortal trellis]

gcd for negative numbers is defined as if you just ignore the negative sign

same for lcm

[proud moth]

Wait for example like gcd(-2,4) would be 2 right but what about lcm(-2,4) wouldn’t that be -2

Is that right?

[plain flame]

No

LCM is positive

By definition

[proud moth]

Ohhh right but like -2 is a multiple of 4 isn’t it??

Oh wait nvm

If it’s negative it could be like negative infinity

I see now so is the lcm 4???

[plain flame]

Yes

[proud moth]

And gcd(-2,4) is 2 right

[plain flame]

Yes

[proud moth]

Oh so a =-2 and b= 4 is a counter example then

.close

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[mortal condor]

I'm going to write my math exam in about 4 days and I'm struggling to understand "Cubic Sequences" would anyone be kind enough to explain it thoroughly? Thanks,

[wary stream]

Have you looked back at notes or looked up resources on that?

[mortal condor]

Both and unable to under stand it

[wary stream]

https://www.youtube.com/watch?v=gIT0YXO1XFY&ab_channel=RadfordMathematics
Here's a video

[mortal condor]

thank you, very much

.close

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[native hemlock]

hi i’ve been trying combinations but cannot seem to get a rule that works :/

only clue i’m onto atm is that they’re not all multiplied but the first digit is added/subtracted due to the 0

[cyan nexus]

I figured it out

Yes, there isnt any multiplication, thanks to the Safe Harabour Vs Dangerous Devils match you can know

They would have scored 0 points then

Wait no 😭😭I was wrong

It doesnt apply for each one

[native hemlock]

i feel like i find a rule

[cyan nexus]

Im a fraud sorry👍

[native hemlock]

and then it doesn’t work for another 😭

[cyan nexus]

Right

[native hemlock]

all good ❤️❤️

<@&286206848099549185>

there doesn’t seem to be a consistent combination…

[charred osprey]

@native hemlock ik it's D

Reason: 9x + 6y + z

[native hemlock]

omg legend

was that just from experience?

[charred osprey]

no

i think im lucky

[native hemlock]

xD

thanks for sharing the luck!!

[charred osprey]

the way i do it is actually getting the equations from 7.9.11 = 92 ; 2.12.5 = 95 ; 1.2.9 = 30 and then set x y z in them and somehow it worked

[native hemlock]

oh amazing!!

tysm, have a good day/night

.close

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[long vault]

Does this have a constant rate of change? I'm trying to figure out if it's linear or not

[slender gull]

Have you done anything?

[long vault]

Not yet, it says to find out if the table is linear or not

[slender gull]

Do you have any idea how you'd proceed?

[long vault]

Not really, no

[slender gull]

Well, can you find the rate of change in y with x, for any two given points?

[long vault]

I have difficulty with it

[slender gull]

Do you atleast know the formula?

[long vault]

Somewhat

I am incredibly forgetful

[slender gull]

"Somewhat"

Alright, whatever you remember.

Try writing it out.

[long vault]

Isn't it two coordinates with a line separating them, or am I confused?

[slender gull]

It is that.

But how'd you write the rate of change?

[lone heart]

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[proper charm]

How do i divide indices with different bases, same exponents

[proper charm]

for example,
6^5 / 5^5

and im talking about simplifying btww

[worn fox]

$\frac{x^a}{y^a} = \left(\frac{x}{y}\right)^a$

[ocean seal]

ΣAC

[proper charm]

ohhhh

thankss

if the question was 30^2 / 5^2 then it will become (6)^2?

[alpine sable]

yeah

[proper charm]

thanks

[nimble fern]

remember to close the channel when you're done!

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[fickle oyster]

from a deck of 52 cards what is the probablity that 2 kings are draw when randomly selecting 4 cards?

[fickle oyster]

isn't this just $\frac{4C2}{52C4}$

[echo socket]

Text out of $$

[ocean seal]

Bedsheat

[fickle oyster]

ohhh i see

[echo socket]

Yeah I think that's the answer

[fickle oyster]

alr thx!

.close

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[summer stratus]

quick question

[summer stratus]

why does every question mention "p is small"

what does it mean

[vale wigeon]

p^2 and higher powers can be thrown out

[summer stratus]

wdym by thrown out

[vale wigeon]

treated as zero

neglected

[summer stratus]

oh alr

ty

i got pi/2 but the ans scheme shows pi/20

[lone heart]

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[robust shoal]

find when the product (r-1)(r+2)^2 is equal to zero and let those numbers be equal to h-5 and solve for h

(a product is zero when one of its terms are equal to zero)

[grave badge]

zeros are given in the equation

take a look at it

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@prime cliff Has your question been resolved?

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[rough blade]

I am bad at math and need help with this
backround info
This is a minecraft minigame where you play hide and seek and you can use taunts. Taunts give xp which increase your level. I want to know the best taunts to use to get max level the quickest. The game lasts 5 minutes and the seeker starts seeking 20 seconds in and you can start using taunts 20 seconds in, so in total I have 280 seconds to use taunts and gain xp while hiding.

Taunt 1 gives 3 xp with a 8 second cooldown
Taunt 2 gives 5 xp with a 15 second cooldown
Taunt 3 gives 6 xp with a 18 second cooldown
Taunt 4 gives 8 xp with a 23 second cooldown
Taunt 5 gives 10 xp with a 30 second cooldown
Taunt 6 gives 12 xp with a 35 second cooldown
Taunt 7 gives 15 xp with a 45 second cooldown
Taunt 8 gives 18 xp with a 54 second cooldown

What is the best taunt to use for the most xp in total during the 280 seconds that i have?

[rough blade]

k thanks

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[velvet musk]

how did they get the coordinates?

[velvet musk]

i mean how did htey get the x values

this is the full question btw, hope it helps

[hidden jackal]

Do you mean the values x = -1.732 and x = 1.732?

[velvet musk]

yes those

how did they get those?

[hidden jackal]

They're finding the intersection between the initial equation y = x^2+4 and the line y = 7

So, set them equal to one another, solve for x

The decimals are just +/- sqrt(3)

[velvet musk]

wait wait set y = x^2+4 equal to 7?

[hidden jackal]

correct

back in algebra, that's how we learned to find where two functions intersect

take the two "y" expressions and set them equal

y1 = x^2+4
y2 = 7

So,
x^2 + 4 = 7

[velvet musk]

ah but what about y=1?

[hidden jackal]

We attempt the same thing, but no solution emerges

So, the line y = 1 does not intersect the equation y = x^2 + 4

y1 = x^2 + 4
y2 = 1

x^2 + 4 = 1
x^2 = -3
No Solution

[velvet musk]

oooh right, can i try a different example and come back if im lost? or do i have to close immediantly?

[hidden jackal]

if it's a different example, probably close this thread and reopen a new help

[velvet musk]

gotcha, lemme try this example myself

ah ok i got it, its also pos and neg because your taking the sq root

gotcha thx man

.close

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[stable portal]

doing this question again and cant help but think its true?

[worn fox]

didnt you come up with a counterexample earlier

[stable portal]

yeah but it feels

iffu

since i proved it too

[lone heart]

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

[worn fox]

proved what?

the statement?

it is not true by your own counter example so you messed up somewhere

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[dapper echo]

I love it when my deniminator suddenly goes from being -(x+1) to -1(x-3)

[uncut torrent]

it goes from 4-(x+1) to -(x-3)

[dapper echo]

but how

[copper delta]

evaluate the denominator

[uncut torrent]

4-(x+1)=4-x-1 = 3-x = -(x+3)

[dapper echo]

oh i see.

.close

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[dry vine]

How to find the surface area

[lapis osprey]

is the 16 cm the height?

@dry vine\

[dry vine]

Yeah

[royal grail]

using Pythagoras, you can first find the length of the diagonal of the base square with side-length 12cm, then Pythagoras again for the diagonal of the triangle, now you have all 3 lengths of the triangle, and you should be able to work out the area of the triangle

[lapis osprey]

^

was about to say that lol

[royal socket]

what a diagram that

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[rotund shoal]

Would this be = -6?

f(2): 4
f(-2): -2
f(-5): -8

[ocean hawk]

looks right

[rotund shoal]

Righto, answer sheet stated something else

just wanted to check I was interpreting the values correctly

thanks

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[warm kestrel]

hi

[warm kestrel]

id like some help with part b

so far ive been able to substitute cosxcotx into the lhs of the equation

said that x = 3x - 50

thus x = 25

but i dont know what else im meant to do now

[granite inlet]

I believe that's it

[warm kestrel]

idk apparently theres 2 other answers

[granite inlet]

Oh I see I guess you could have where cos(x)=0 be a solution

Certainly 0=0 is true

[warm kestrel]

wait how

did you get cos(x)=0

[granite inlet]

When you substitute you get

$/cos(x) /cot(x)= /cos(x) /cot(3x-50)$

[ocean seal]

Blue Guilmon

[granite inlet]

Oh they aren't functions in this tex

[pseudo ice]

$\cos(x)$

[granite inlet]

But you get what I mean just disregard the /'s

[ocean seal]

@pseudo ice

[granite inlet]

Oh soz on my phone xD

[warm kestrel]

ah ok

no no dont worry its ok

[warm kestrel]

ok so

[granite inlet]

Try solving for the zeros of this instead

[warm kestrel]

where x = 0?

why

[granite inlet]

$\cos(x) \cot(x) - \cos(x) \cot(3x-50)=0$

[ocean seal]

Blue Guilmon

[warm kestrel]

o

[granite inlet]

Then it becomes more obvious why cos(x)=0 is a solution

[warm kestrel]

but then dont we also have

cot(3x - 50) = 0

[granite inlet]

Factored you get $\cos(x)[\cot(x)-\cot(3x-50)]=0$

[ocean seal]

Blue Guilmon

[warm kestrel]

ahh

okay

so then x = 25 comes from inside the brackets

and then

[granite inlet]

Yup

[warm kestrel]

cos x = 0

[granite inlet]

Because if $a \cdot b =0$ then either $a=0$ or $b=0$ in algebra

[ocean seal]

Blue Guilmon

[warm kestrel]

so then x = 90

[granite inlet]

Yup

[warm kestrel]

wait

theres another solution of x = 115

[granite inlet]

So

3?

[warm kestrel]

3 solutions?

ya

idk where the 115 comes form

from

90+25

aha idk

i think

[granite inlet]

Hmm

[warm kestrel]

the cot(x) = cot(3x - 50)

[granite inlet]

Try multiplying everything through by $\sin(x)$

[ocean seal]

Blue Guilmon

[granite inlet]

Because $\csc(x)=\frac{1}{\sin(x)}$ on the left and so you get $1-\sin^2(x)$ on the left which using pythagorean's is $\cos^2(x)$

[ocean seal]

Blue Guilmon

[warm kestrel]

isnt it cos(x) ?

outside the bracket

[granite inlet]

Then solve the zeros of $\cos^2(x)-\cos(x)\sin(x)\cot(3x-50)=0$ yields an equation $1-\sin(x)\cot(3x-50)=0$

[ocean seal]

Blue Guilmon

[granite inlet]

I may be overcomplicating it tho

[warm kestrel]

it could work but this question is technically meant to take 5 minutes lol

[granite inlet]

Yeah it's a bit tricky but not hard

[warm kestrel]

the answers to get 115 say this

if that makes any sense to you

[granite inlet]

Yes it does

I see

Yeah cotangent had a period of pi or just 180

Has*

So $\cot(x)=\cot(x+ \pi)$

[ocean seal]

Blue Guilmon

[granite inlet]

Substitute that into your original cotangent equation on the left

Except maybe 180 instead of pi if they're using degrees

[warm kestrel]

wait what

ya its degrees but

how is cot(x) = cot(x+180)

[granite inlet]

Cotangent has a period of 180

It repeats every 180

[warm kestrel]

oo

yea

okay ya that makes sense

so then cot(x + 180) = cot(3x - 50)

[granite inlet]

Yup

So x+180=3x-50

[warm kestrel]

and then x = 115

i see i see

[granite inlet]

Which fits in our bounds

In general

[warm kestrel]

mhm

[granite inlet]

You probably wanna write a general solution to these

[warm kestrel]

general solution?

[granite inlet]

That would've saved us a lot of trouble haha

Yeah trigonometric equations usually have a solution that is periodic

Meaning they happen on regular intervals because the functions themselves are periodic

[warm kestrel]

oh true

[granite inlet]

So if you are working with specific bounds

Write up a general unrestricted solution first

Then just plug in numbers until you get everything in your bounds

[warm kestrel]

yeah that makes sense

well tysm for helping !

[granite inlet]

Np

Sorry on my phone it's slow xD

[warm kestrel]

nono its okay you were great

thank you

.close

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[merry plinth]

$$
\int \frac{dx}{a^2\sin^2x+b^2\cos^2x}
Given ab not equal to zero.

This can be solved by dividing by \cos^2x quite easily. This is also the solution in the solution manual is to divide by \cos^2x then use u-sub. for tanx. One question though. How come it is valid to divide by cos^2x? Wouldn't it sometimes be zero for the interval over which the integral is being calculated?
$$

[ocean seal]

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

[surreal meadow]

$$
\int \frac{dx}{a^2\sin^2x+b^2\cos^2x}
$$
Given $ab$ not equal to zero.

This can be solved by dividing by $\cos^2x$ quite easily. This is also the solution in the solution manual is to divide by $\cos^2x$ then use u-sub. for $\tan x$. One question though. How come it is valid to divide by $\cos^2 x$? Wouldn't it sometimes be zero for the interval over which the integral is being calculated?

[ocean seal]

maximo

[surreal meadow]

what do you mean by "divide by cos^2 x", can you show what they did? @merry plinth

[alpine sable]

where it is 0 has measure 0 or sth

[surreal meadow]

that's not enough jester

you can't integrate 1/(x^2) from -1 to 1, even though 1/(x^2) is only udnefined for a singleton

[merry plinth]

Sure!
$$
\int \frac{dx}{a^2\sin^2x+b^2\cos^2x} =
\int \frac{sec^2x dx}{a^2\tan^2x+b^2}
Let u = tanx
du=sec^2x
\int \frac{du}{a^2u^2+b^2}
$$

[ocean seal]

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

[merry plinth]

Sorry, Idk how to properly format for the bot

[surreal meadow]

,tex
\begin{align}
\int \frac{dx}{a^2\sin^2x+b^2\cos^2x} = \
\int \frac{\sec^2x dx}{a^2\tan^2x+b^2}\
\text{Let }u = \tan x\
du=\sec^2x\
\int \frac{du}{a^2u^2+b^2}\
\end{align}

[granite inlet]

Feels like that was the culmination of almost all of real analysis until it wasn't

[ocean seal]

maximo

[surreal meadow]

they didn't divide by cos^2(x)

[merry plinth]

There we go. Except im missing a dx lol

[surreal meadow]

i get what you're saying though

what are the bounds for the integral?

[merry plinth]

It is an indefinite integral

[surreal meadow]

then you would just say this works only when cos^2(x) =/= 0

same way it only works when a^2sin^2(x) + b^2cos^2(x) =/= 0

[merry plinth]

i mean but the author doesnt state anything about cos^2x not being zero

so would solving it that way not be a full solution?

[surreal meadow]

do they state anything about a^2sin^2(x) + b^2cos^2(x) not being 0?

[surreal meadow]

it would still be a full solution, as far as your class is concerned

[merry plinth]

uhh, i dont think its possible for it to be zero given ab not being zero

tho maybe it can be for negative values of a and b

[surreal meadow]

it's implicit that the denominator is nonzero

[merry plinth]

yes

[surreal meadow]

via the same handwaving we just take cos(x) =/= 0

and it may even converge in that case still

but that's not what this problem is about

[merry plinth]

also it looks like a=-1 and b=0 results in a function that is sometimes zero so yes sometimes the denominator is zero for certain choices of a and b

[surreal meadow]

ab is nonzero

[merry plinth]

yes

[surreal meadow]

so those choices aren't allowed

the denominator is never zero

the solution given is complete as far as your class should be concerned. is this for a calc 2 or similar class?

[merry plinth]

a=-1 and b=1 means ab=-1 tho

[surreal meadow]

a and b are squared in the denominator

they will both be positive

as will sin^2 and cos^2

[merry plinth]

its not a class im just doing problems provided in a calculus book that ive been slowing working through

[surreal meadow]

then chances are the calculus book isn't concerned with the case for cos^2(x) = 0

[merry plinth]

oh wait yeh ur right so the denominator is never zero for any choices of a and b

[surreal meadow]

whatever you get for the answer, plug in a value that would make cos(x) = 0

[merry plinth]

im confused why it would not be concerned for choices of cos^2x = 0 though?

[surreal meadow]

because calculus books at these levels are more about integration techniques rather than being rigorous

have you seen a rigorous proof for u-subtitution working?

if you want to justify what happened for values of x making cos^2(x) = 0
note that sin^2(x) = sin^2(x) * cos^2(x) / cos^2(x), for values of x such that cos^2(x) is nonzero

for those values of x, though, we can still extend the function sin^2(x) * cos^2(x) / cos^2(x) naturally to the value of sin^2(x)

[merry plinth]

the book did in fact cover a moderately rigorous proof of u-sub

[surreal meadow]

moderately rigorous should be the key term for you

[merry plinth]

okay so taking a definite integral from pi/2 to pi for example you would end up with zero in the denominator of the tan function tho?

[surreal meadow]

you would just not consider that point

[merry plinth]

i mean that is it graphed on desmos so i dont see why it wouldnt be considered

[surreal meadow]

because there is a natural extension for it

what is that graph for

[surreal meadow]

yes that's still well defined

[merry plinth]

its just the graph of the function with some choices of a and b

[surreal meadow]

i don't see why that point wouldn't be considered there

graph sec^2(x) / (tan^2(x) + 1)

https://www.desmos.com/calculator/t20eubtivw

[merry plinth]

oh, they produce the same graph

[surreal meadow]

yes, they are equivalent, but as you said one is not defined for specific values of x

but they have a natural extension to the graph where it's well defined

[merry plinth]

that's what i would have thought

[surreal meadow]

in particular, the evaluation at those points given by the former function makes our latter function continuous

[merry plinth]

im confused how that is possible

[surreal meadow]

how what is possible

consider the graph of 1 v.s. the graph of x/x

one is defined at x = 0, the other is not

but it makes sense to extend the graph of x/x to be 1 at x=0

[merry plinth]

how the sec^2x/(4tan^2x+16) is continuous

[surreal meadow]

it is not

desmos just simplifies and assumes the former function

[merry plinth]

oh okay

[surreal meadow]

because it's the most natural extension for it

[merry plinth]

i guess there is still a hole then at pi/2?

[surreal meadow]

yes

but

what is the integral at a single point?

think of a riemann sum

and think of what the with of the rectangle would be

[merry plinth]

its zero. or rather can be ignored since it does not contribute

[surreal meadow]

yes, and the reason why it can be ignored is because it is a hole

[merry plinth]

i guess the point is since the area being excluded by cos^2x = 0 is only at single points that it doesnt impact the sum at all?

[surreal meadow]

it's more that

we can include those points with a natural extension of the function

hence why they are holes rather than jumps or asymptotes

[merry plinth]

when you say you can include those points with a natural extension do you just mean redefining the function to include the points where there are holes at those points?

[surreal meadow]

yes, kind of like $f:\mathbb{R} - {0} \to \mathbb{R}, f(x) =\frac{x}{x}$ and the extension $g:\mathbb{R}\to\mathbb{R}$ given by $$g(x)= \begin{cases} f(x) & x \ne 0 \ 1 & x = 0\end{cases}$$

[ocean seal]

maximo

[surreal meadow]

note that g(x) is just the function g(x) = 1

this is a very simple case of such an extension, and not all functions have extensions like this

but in your case, we do, so we can just ignore the missing points knowing there's a nice inclusion

[merry plinth]

well its also possible since the right and left limit approach the same point

[surreal meadow]

yes

that was another way to look at it

let f be the integrand for your integral

then we just consider $$\lim_{\varepsilon\to 0} \int_0^{\frac{\pi}{2} - \varepsilon} f(x), dx + \int_{\frac{\pi}{2} + \varepsilon}^\pi f(x), dx$$

[ocean seal]

maximo

[merry plinth]

that would be the same as just taking:
$$\int_0^{\pi}f(x)dx$$
right?

[ocean seal]

Lily.

[lone heart]

@merry plinth Has your question been resolved?

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[rapid scaffold]

im making my final SAT revision but some word problems are confusing me, particularly this

[buoyant kayak]

have you tried anything yet?

[alpine sable]

b = beef
c = chicken

[buoyant kayak]

@rapid scaffold

[rapid scaffold]

yeah

im trying to do it mentally quick according to exam setting but i can't decipher it much

i suck with word problems

[buoyant kayak]

when you see "equal to", what do you think you should do?

if b and c represent the prices and they are equal

[rapid scaffold]

make x subject

then plug into other equation

[buoyant kayak]

not quite

[rapid scaffold]

that's what I'd do in a regular situation

[buoyant kayak]

you have equations for b and c and you want to find when they are equal

so if you were to set b equal to c, could you find a value for x?

[rapid scaffold]

now that makes sense

im used to long-form questions so word problems are new to me

[buoyant kayak]

and remember, when you do set the equations equal to each other and solve for x, x represents the number of weeks

and you're asked to find a price

[rapid scaffold]

something like this is a bit easier

since i do the inequality normally

[rapid scaffold]

then plug x back into either equation

actually both

value should be the same then

[buoyant kayak]

correct

[rapid scaffold]

this should be 4x - 3x - 3 <= -5 correct?

but then it gives x <= - 2

[buoyant kayak]

right

so all solutions less than or equal to -2 would be valid solutions

and only one of those options isn't

[rapid scaffold]

-1 is wrong then

[buoyant kayak]

wrong in what sense

[rapid scaffold]

it doesn't fit the inequality

i should've said invalid

[buoyant kayak]

yes, correct

[pseudo ice]

Could you please post this in an available channel in the Math Help Available section? [bear in mind that #❓how-to-get-help isn't telling you to take channel 0 specifically, just one of the ones that are unoccupied (so have no names on them)]

[meager lark]

Ok thank you

[lone heart]

@rapid scaffold Has your question been resolved?

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[keen orbit]

oh, lmao, I didn't see the unconnected dot in A

nvm

.close

[lone heart]

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[rotund shoal]

For this, why is it a relation

I understand why it is one to many now

I used the vertical line test

is it assumed that the middle section in that graph is exactly parallel to the y axis?

[neon scarab]

Even if it is true, there is no way for student to guess teacher’s intention just by looking at the picture. 😂

[rotund shoal]

Okay, but my original question is why is it a relation?

Let me confirm, the red pen is the correct answer

[granite badger]

looks pretty function to me

[rotund shoal]

Thats what I thought too... the answer sheet says it is a relation

thus, I was asking would I have had to assume that this part in particular was parallel to the y axis

Because that would be the only way where a line of x would pass through the graph twice

[neon scarab]

The graph shows a (infinite) collections of ordered pairs (which is infinite since the graph is continuous). And such collection is a relation if not a function.

[rotund shoal]

Could you elaborate on that, I don't fully get what you mean

Right now, the only method I have of determining if a graph is a function or relation is by inspection

and thats purely through a vertical line test

[granite badger]

this is a point of inflection, more specifically a point of infinite steepness

[neon scarab]

Quick googling on the exact definition on "relation" first.

A relation between two sets is a collection of ordered pairs containing one object from each set.

The graph you are having joins a bunch of (x, y), and the graph tells whether and how points should be joined. This "rule" itself is a relation.

[rotund shoal]

So are you saying for a function, it needs to be one to one?

[granite badger]

it should be a function, flipping through my notes it seems like it’s a graph of -x^(1/3)

[neon scarab]

Some other googling.

The difference between a relation and a function is that a relationship can have many outputs for a single input, but a function has a single input for a single output.

So loosely speaking, a 1-to-1 (and also many-to-1) "rule" is a function, and all others are relations.

[rotund shoal]

Right, I thought so too. In that case would the correct answer also be one to one?

Additionally, for this question. Would it be a relation and many to one?

Because for when y=0

you have 3 seperate roots of x

[neon scarab]

Yes for (g) your answer key or your teacher is definitely wrong. 😂

[proven leaf]

What is the "M-O", "O-O", and "M-M" notation begin used here?

[rotund shoal]

Me being too lazy to write many to one, one to one etc

[neon scarab]

It’s ok, in higher lv math course you won’t be needing to write these terms that often.

[proven leaf]

ah ok ty

[rotund shoal]

Anyway, I think it should be fine. A lot of the ones that I got wrong I believe it may be the answer sheet error

Thanks for the help

.close

[lone heart]

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[wispy rampart]

I don't need help with the whole question i just have a question for a specic part of the equation. When working what the quadratic equation i noticed that depending on which value of x i used i would either get the right answer or wrong answer. And im wondering does this mean that i will only get the right answer if i use an a value of x equal to or less than 3? My question wasn't answered before it got closed so I'm asking again if thats fine

[vapid shuttle]

that is fine to ask again, btw next time just react to the bot's message with a ❌ when it asks if your question is resolved, and that will keep your channel from being closed

Could you be more specific what your question is?

[wispy rampart]

basically I am trying to solve A. When I solve for A using the points (2,4) I get the correct answer of -1/2. But when i used the points (6,0) I get a wrong answer of 0. So i am wondering why that is the case

[vapid shuttle]

Could you please tell me what "A" is?

What is the original question that you are looking at

[wispy rampart]

Ok so i am trying to find the equation y=ax^2 + bx which is a quadratic. (C=O) And I am trying to solve it by using the turning point and differentiating the equation into 2ax +b = 0 (since dy/dx at turning point is 0). I then make b = -6a. Then substitute in b=-6a into the original y=ax^2 + bx quadratic

this was the og question

[vapid shuttle]

and you are trying to find the equation of that quadratic?

[wispy rampart]

yeah

[vapid shuttle]

Okay gotcha

[wispy rampart]

I know that theres another method to do it. I'm just trying to figure out why the method im using spits out two different answers depending on whether i use points (6,0) or (2,4)

which my working in the image i sent kinda shows

<@&286206848099549185>

[lone heart]

@wispy rampart Has your question been resolved?

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[acoustic mirage]

f(x) = 3 + (x-2)^(2/3) in [1,3] , is rolle's theorem applicable on this function ?

[acoustic mirage]

it is continous and it is diffrentiable when x is not equal to 2

so rolle's theorem should be valid right ?

[tacit arch]

Do you know the requirements of Rolle's theorem

[surreal meadow]

what makes this not differentiable at x = 2

[acoustic mirage]

yea , it needs to be continous in [1,3] and diffrentiable in (1,3)

this is it right ?

[tacit arch]

Your function is vague.

What's the exponent on (x-2)

[acoustic mirage]

f'(x) was coming as infinity so hence it it not diffrentiable ig

[acoustic mirage]

(2/3)

[surreal meadow]

oh i thought the exponent was a 2

[tacit arch]

Does this answer your question

[acoustic mirage]

yea sorry mb , i edited it

[acoustic mirage]

does what ans my ques ??

[tacit arch]

.

[acoustic mirage]

but its only non diffrentiable at one point ?

for values greater than 2 it would be diffrentiable

[tacit arch]

"Differentiable on (a, b)" means differentiable at ALL points a<x<b

[acoustic mirage]

oh i wasn't aware of that 💀

ok so ig this clears my doubt thanks

.close

[lone heart]

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[livid crypt]

This is my work so far can anyone help steer me in the correct direction?

I'm not sure if im doing it right

for A

I'm trying to work out the proof for k+1 but im kind aconfused

[vale wigeon]

better to put question marks above the <'s for P(k+1) and the stuff that follows

$(k+1)!^2 \neq (k+1)^2!$ btw

[ocean seal]

Ann (glomed)

[vale wigeon]

so you have $(2^{k+1} (k+1)!)^2 \overset?< (2k+2)! (k+2)$

[ocean seal]

Ann (glomed)

[vale wigeon]

from (2^k * k!)^2 < (2k)! (k+1) check the factor by which each side grows

on the left that will be (2(k+1))^2 and on the right it will be (2k+2)(2k+1)(k+2)/(k+1), or 2(2k+1)(k+2)

compare those

glossing over some details here jic

[lone heart]

@livid crypt Has your question been resolved?

[livid crypt]

I worked on it and got a similar answer to yours thank you so much!!!

[lone heart]

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[livid crypt]

Can someone check this work for my problem?

[lone heart]

@livid crypt Has your question been resolved?

[lament forge]

``for any $y \in \mathcal{E} \times \mathcal{E}$ there exist $k \in \mathbb{N}$ such that $f(k) = y$" \ what about $(2,4)$?

[ocean seal]

bee [it/its]

[lament forge]

@livid crypt

[livid crypt]

What do you mean what about (2,4)?

[lament forge]

if y = (2,4) then what's the value of k such that f(k) = y?

it looks like your proof claims that it's y/2, so in this case (2,4)/2, but i don't know what that means

[livid crypt]

This is my friends work and im trying to make sense of it for my understanding

im lost which is why Im looking for clarification

but are you saying that f is not onto?

[lament forge]

yeah the function they defined in c isn't onto

[livid crypt]

so c is wrong?

is a and b correct?

[lament forge]

i think their answers to a and b are correct

[livid crypt]

okay thank you so much

would C not be countable?

[lament forge]

no E x E is definitely countable

they just took an approach to proving that that doesn't work

[livid crypt]

oh okay thank you for helping anyways

I appreciate it

unless u have a proof that works I can try to find it on my own

so you can end this channel so someone else can ask a question

[lone heart]

@livid crypt Has your question been resolved?

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[native nebula]

help

[native nebula]

example 21

what is that with 2! ways?

why that is written?

why is that required that in how many ways vowels can be arranged?

[rose sigil]

it's like... if you look at all the permutations of GARDEN

all 720 of them

you could make 360 pairs of them, where two are paired if they just have A and E swapped

so NARDEG and NERDAG are "paired"

for example

and only one thing in each of the pairs is valid (the one that has A before E)

so half of the total arrangements are valid

[native nebula]

but 2! has nothing to do with it

[rose sigil]

2! = 2

[native nebula]

yeah

[rose sigil]

it just coincidentally looks like 2! has nothing to do with it because 2! = 2

what if the word was GARDEO?

[native nebula]

the encircled one

this line is irrelevant

[rose sigil]

it's not irrelevant lol

if there were 3 vowels, the next line would instead say in 1/3! of the 6! arrangements blah blah blah

instead of this which is basically: in 1/2! of the 6! arrangements

[native nebula]

ok

got you 😁

thank you layla

.close

[lone heart]

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[rose sigil]

np!

[lone heart]

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[deep crater]

For this question, it says to prove it, but im finding that it's false, is this correct?

(Sin(x) / 1 + sin(x)) - (Sin(x) / 1 - sin(x)) = 2tan^2(x)

Btw, I got 2tan(x) = 2tan^2(x) in the end

[hushed locust]

as it stands that's not true although it does equal -2tan^2x

[lone heart]

@deep crater Has your question been resolved?

[deep crater]

So what would happen then?

[hushed locust]

i would ask your teacher about that since you can't really prove a false statement

try to work to prove that it equals -2tan^2x in the meantime

[deep crater]

Hmm

[deep crater]

Did you get 2tanx tho?

[hushed locust]

no, that's not correct

[deep crater]

What did you get then?

[hushed locust]

as i said, i got -2tan^2(x)

[deep crater]

Can you show how?

[hushed locust]

try multiplying each by 1 (same on top and bottom) to make a difference of squares

[deep crater]

Yeah I'm still confused

[hushed locust]

so we can always multiply something by 1, and 1 can be any fraction as long as the numerator = denominator

the denominators on each fraction look like a difference of squares, so try converting each fraction to a common base using that

[deep crater]

(1+sin^2(x))(1-sin^2(x))?

[hushed locust]

ok so let's start with $\frac{\sin(x)}{1+\sin(x)}$. What happens when we multiply by $\frac{1-\sin(x)}{1-\sin(x)}$?

[ocean seal]

cloud

[deep crater]

The bottom and top get that value

[hushed locust]

what do you mean?

[deep crater]

They get those value added once multiplied

[hushed locust]

so what does your first fraction look like now?

[lone heart]

@deep crater Has your question been resolved?

[lone heart]

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[twilit hill]

Why doesnt this function have an absolute maximum? Is it due to the interval being open on one end?

[uneven jewel]

yea that's the gist of it

f(x) has a least upper bound of 4 on that interval although it cannot actually attain that bound

for an absolute max it needs to be the least upper bound and attainable

you can inch up as close to x=2 as you want but can't reach it, at the same time f(x) inches towards 4

[twilit hill]

Thanks

[uneven jewel]

np

[twilit hill]

.close

[lone heart]

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[dusky chasm]

can someone explain to me how the answer is not 1/3

[dusky chasm]

all i did was find the integral between 0 and 1 of xf(x)dx

[vale wigeon]

do you have work for how you calculated the integral?

you most likely screwed up at some point in your calculations

oh right

no ok i can see it already

@dusky chasm the support of f is [0,2] NOT [0,1]

you neglected half of the interval over which you were meant to integrate

[dusky chasm]

wait so if i get a question like this

i should consider what cause idk where you got that 0 to 2

domain

and i think it still gives me the wrong answer cause the answer is one

[vale wigeon]

look at the defn of f(x)

do you see that it is nonzero from x=0 to x=2

[dusky chasm]

yes

[vale wigeon]

yeah

IN GENERAL the expectation of a random variable is the integral of x f(x) over the entire real line

ie from -∞ to +∞

[dusky chasm]

yes

[vale wigeon]

its just common practice to immediately clip off the parts where the density is 0

since, yknow.

it's 0

[vale wigeon]

have you redone your integral calculations?

[dusky chasm]

it gives me the wrong answer still

[vale wigeon]

what answer do you get?

[dusky chasm]

even when i change the bounds

[vale wigeon]

what answer do you get?

[dusky chasm]

i got 8/3

[vale wigeon]

and again can you show your work?