help-0

[stone barn]

It*

Damn I should be screwed for my exams if I can’t even understand this

[mossy swift]

but like i thought we already went over this :V

the total in the sample space without restriction is quite literally 9C5

uhhh

wait so

[agile sundial]

i think the examiner didn't consider that the fact that there are repeated letters

[mossy swift]

consider AAB as heads and tails instead where A -> heads and B -> Tails

the sample space of all 2-element tuples are: (head, tail), (head, head), and (tail, head)

clearly (head, tail) and (tail, head) are different cuz (head,tail) -> head first then tail and (tail,head) -> tail first then head

[stone barn]

The word is ACTIVATED, 1 and 2 is used to mark the first and 2nd A and T
In A(1),C,T(1),I,V,A(2),T(2),E,D you have 2T’s and 2A’s so if we use total selection 9C5 then A(1),C,T(1),I,V and A(2),C,T(2),I,V is gonna be considered as 2 selections when in fact they are both ACTIV

This is what I can condense my question as

[mossy swift]

and my response is with your previous example:

AAB if you want to find all the two element tuples (this is different from set so something like AB and BA are different) is quite literally 3C2 because:

1. AB
2. BA
3. AA

[stone barn]

So if we had 10 E’s like EEEEEEEEEE and we gotta randomly select 5 then 10C5 would be answer tuple wise for the total number of combinations and if it is in set then it would only be one answer EEEEE?

[mossy swift]

yeah there's only 1 tuple lol

namely: (E,E,E,E,E)

not 10C5 of them

[stone barn]

so repetition does matter?

[stone barn]

Then what abt this

If the first ACTIV is same as second ACTIV wouldn’t the total number of selection at least be 1 less than 9C5

[mossy swift]

it seems like the repetition only matters when you have literally an element with full-on repeats

consider AAAC or AAAAAC or AAAAAAAAAAAAAC

then nC(n-1) works to pick 2 element tuples

AAAC

1. ACA
   2)AAA
2. CAA
3. AAC

which checks out with 4C3 = 4

anyway i'm going to eat

[stone barn]

so in a word like RABBIT I gotta select 2 letters it would be 6C2?

Acc fk it

in exam they use 9C5 without caring about repetition I’ll follow it

Anyway rn I’m pursuing to pass not find out the truth

Thanks for putting up with this

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[plain schooner]

i think i know how to do this but just incase

i think it is 64/361

<@&286206848099549185>

[ivory fern]

It is not 8/19 times 8/19

Because once you draw a red one, there will only be 7 red marbles

[plain schooner]

OHHH

YEAH I FORGOT

[ivory fern]

And 18 marbles in total

[plain schooner]

but wouldnt the second time it would be 7/18

so 56/342? @ivory fern

[ivory fern]

Yes

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[rich walrus]

What does the first equation represent?

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[left wharf]

Context?

[rich walrus]

you're supposed to find a,b real numbers so that the system has only 1 solution

@left wharf hey?

[left wharf]

maybe you could try writing z=x+yi and then you'll get a system of 3 equations

it won't be a linear system of equations but maybe you'll figure something out

[rich walrus]

3 equations?

[left wharf]

yes because z is zero iff x=y=0

[rich walrus]

well I plugged it in the second one which is a bit simpler and I got y = x , can I use that in the first equation after I plug in z = x + iy and then swap y with x

[rich walrus]

@left wharf okay sorry for taking so long or if im interrupting anything but I got an equation for a circle:

so I tried with a graphing calculator and if a = 0, b can be anything and it will be a single solution

but I wouldn't know how to figure that out on my own..

that's how I got it and below is a sketch of the solutions where a = 0 and b is real\0

anyone?

<@&286206848099549185>

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[low kestrel]

How do you linearise data relating to Temperature and Time for a system following Newton's Law of Cooling.

[jagged cobalt]

T-T_amb=(T0-T_amb)e^{kt}
log(T-T_amb)=log(T0-T_amb)+kt

[low kestrel]

Sorry, could you explain please?

[jagged cobalt]

to make an exponential function into a linear form, we take logs

its
$$T=T_a+(T_0-T_a)e^{kt}$$
if we want to make it linear the right side needs to be a product (so we can split the log), not a sum so move $T_a$ across
$$T-T_a=(T_0-T_a)e^{kt}$$
$$log(T-T_a)=log\left[(T_0-T_a)e^{kt}\right]$$
$$=log(T_0-T_a)+log\left(e^{kt}\right)$$
$$=log(T_0-T_a)+kt$$

Hence:
$$log(T-T_a)=log(T_0-T_a)+kt$$

[ocean seal]

AℤØ

[jagged cobalt]

y intercept is log(T_0-T_a), slope is k, y is log(T-T_a), x is t

y=mx+c

[low kestrel]

why are you logging it?

[jagged cobalt]

to bring kt down from the exponent

we want it to be linear so has to be of the form y=mx+c

[low kestrel]

oh ok

itd be ln though wouldnt it?

[jagged cobalt]

im using log for base e yeah

[low kestrel]

ok thanks

[jagged cobalt]

you can use other bases, only difference will be the slope becomes klog(e) instead of just k

[low kestrel]

right

but when you log, one of the T_0 -T_a, dissappears, why?

[jagged cobalt]

where did it disappear?

theres only one T_0-T_a

[low kestrel]

oh sorry i meant the T-T_a

[jagged cobalt]

its still there, i just didnt feel the need to keep writing on the left side of the equation

i bring it back on the last line

[low kestrel]

ah ok

reposting for ease

what do the values mean? we just started this so im not quite sure

like T and T_a

[jagged cobalt]

T is the temperature at time t
T_a is the ambient temperature (temp of surroundings)
T_0 is the initial temperature
k is just a constant, changes based on the object

[low kestrel]

right

they've given us a table of values, would you mind jsut helping me out with it as I do it?

[jagged cobalt]

sure

[low kestrel]

ok thanks

[jagged cobalt]

is the ambient temperature known?

[low kestrel]

24 degrees

[jagged cobalt]

alrighty, i assume you can probably use excel or equivalent for this

[low kestrel]

yeah were using google sheets

but you cant just plug in the values as is right

to the linearised formula

[jagged cobalt]

unless you know k, no

[low kestrel]

so would we solve for k?

we have all other values

[jagged cobalt]

if you plot a graph of log(T-24) against t, you should be able to find it

i dont use google sheets but i assume it can generate lines of best fit and their equations

[low kestrel]

yeah

im just going to paste everything into a doc so i can read it easily

i have the values they gave us in sheets, i just get log(T-24) and make T correlate with each time frame

[jagged cobalt]

yup

first one for example is point (0, log(49))

which is log(T_0-T_a) in our formula as desired

[low kestrel]

but then how does kt affect it

also if you plug in the values starting with 0, then you cant do it right

because log cant have a value of 0

<@&286206848099549185>

[jagged cobalt]

where is the log taking a value of 0?

there isnt one

[low kestrel]

the first value of T is 0

[jagged cobalt]

the first value of T is 73

t is time, its first value is 0

[low kestrel]

oh i see my mistake

what do the linearised values really mean

I have some but I don't think they're correct

[jagged cobalt]

when you linearise it its easier to extrapolate future values from your data

[low kestrel]

right

these are the values I got after linearising them

They're not correct though

For reference, this is what it should be

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[low kestrel]

<@&286206848099549185>

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[pseudo seal]

what is the process I have to do to get the other coordinate?

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[steep rain]

any point on the unit circle must has one property in common

you know the distance from point to origin, so you know coord

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[delicate wave]

Central Limit Theorem for Sample Means

Hello, could someone help me and show me how I'm supposed to do this? My teacher wasn't very good at showing me how to do these problems so I'm a bit confused.

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[dull pilot]

Can someone help how to prove theta = pi/2?

It's apparently correct but I don't understand how to algebraically prove it

[dull pilot]

My logic is: in order to get positive r I have to multiply both sides by -1

and I'll always end up with a negative* -3?

Unless when you go cos(-theta) to cos(theta) do you also change the number in front of cos to fit the argument?

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[charred flint]

I don't think you need the -r part so what you did works out

like being symmetric across this theta=pi/2 line still has positive r

[dull pilot]

Logically what you say makes sense I agree

But the book says above Replacing (r, theta) with (-r, -theta)

Why would it say that?? ?

[charred flint]

oh huh, it works for both (-r,-0) and (r,-0), I'd have to think about it

it's kinda weird because the functions aren't really equal, but when you graph them fully from -2pi to 2pi they draw the same graph, just one's in reverse

[dull pilot]

r = 3 cos(theta/2)?

I see that, I only got the answer right because I desmos'd it : ) - but if it was a prove algerbraically I'm sol

[charred flint]

yea I'm playing with desmos to see lol

like you're allowed to flip it because it's on the whole domain so it works out, but the solution doesn't mention that

it asks you to graph it though so I guess that's expected

like sin or cos on a full domain can have their sign flipped to give the same shape, it just draws it in an offset way

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[lapis swan]

hi

[lapis swan]

What are the answers for 1-3?

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@lapis swan Has your question been resolved?

[remote panther]

!noans

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The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[remote panther]

What have you tried yourself, where are you stuck?

[lapis swan]

Well

[lapis swan]

I got the answer to 1 and 3

but I just understand NOTHING at all about 2

[remote panther]

Okay because you asked for the answers 1-3

I'll look at exercise 2 the

[lapis swan]

I just solved it

I got C and 3/4's for 3

[remote panther]

But number 2 also looks solved

I'm a little confused to where you need help

[lapis swan]

thats what I tried

I'm not sure if its correct

[lapis swan]

that's why I'm asking

[remote panther]

Not sure about 2d but the rest seems good to me

[lapis swan]

fair

[lapis swan]

what do u think 2d could be

[remote panther]

That kind of depends on what definition of limit you use. There are newer versions and older versions. Hence, I can't give you an answer on that.
I think with the newer version it would be that your function has a limit everywhere. The old one I think would have a problem with x=-3
But as I said I'm not too sure. I should also go to sleep because I'm tired haha

[lapis swan]

ok

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[autumn matrix]

my desmos isnt working anymore could someone help me out

would greatly appreciate it just trynna finish my lesson and head to bed

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[harsh lion]

In a quadrangular cone $P-ABCD$,the base $ABCD$ is square, $AB=4,PC=PD=3,\angle PCA=45^\circ$,the area of $\triangle PBC$ is

[ocean seal]

e_waste

[lone heart]

Please don't occupy multiple help channels.

[harsh lion]

how do I solve this?

heeelp

<@&286206848099549185>

[queen aspen]

can i get a bilateral view of the problem?

@harsh lion

[worn fox]

Can you stop

[harsh lion]

no it requires you to draw your own

but lemme see

I dont have bilateral view

but I think this is enough

[harsh lion]

if you're willing to provide detailed steps I'm sure he's happy to stop

[worn fox]

He's spamming channels with that message

[harsh lion]

oh really...

damn it

really? nobody?

[lunar mica]

all my brain power went to trying to understand wtf that dude meant by bilateral view 💀

[harsh lion]

Well this may help

I have a series of problem and I'm really wanna someone to help me with solid geometry

[exotic canopy]

you get that AC is 4sqrt(2)

from the square

[exotic canopy]

and BD aswell

[harsh lion]

yeah

[exotic canopy]

since PDC has two equal sides, it's isosceles

[harsh lion]

but I know 0 angles in PDC

[exotic canopy]

you know all the sides

[harsh lion]

oh yep I know

[exotic canopy]

actually idk how that i helpful lol

[harsh lion]

so how do I transform PDC into PCB

[exotic canopy]

but you need to find PB and you can do it from triangle PDB

[exotic canopy]

you don't

[exotic canopy]

you have two sides and the angle

[harsh lion]

but idk how can I get angle PDB

[exotic canopy]

by symmetry angle PDB has to be 45 aswell

[harsh lion]

if I can get that I can get PB

oh I see

then it is solved

[exotic canopy]

yea

it's a lot of angle and line chasing really

[harsh lion]

You just need to learn to draw diagram

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[velvet leaf]

Hellpp how to this step by step

[velvet leaf]

Do*

[cinder tundra]

What did you try?

[lone heart]

@velvet leaf Has your question been resolved?

[velvet leaf]

Wait let me send the solution real quick

Let me writeee it

[half arch]

Btw sorry to interrupt I also have a integration question can I ask

[velvet leaf]

@cinder tundra

[velvet leaf]

What integration is it

[half arch]

D

[velvet leaf]

What the fuck is that

[velvet leaf]

@cinder tundra 👉🏻👈🏻

[cinder tundra]

You did good but not finished

@velvet leaf

[velvet leaf]

Whats the next step

[cinder tundra]

You have tan(arctan(something))

You can solve that

[velvet leaf]

Bro what the

This part?

@cinder tundra

[cinder tundra]

Yes

Looks correct to me

[velvet leaf]

Thank you bro

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[lean sun]

I'm currently doing a calculus paper for my classwork, the question word for word is: "For each following non-routine functions, calculate first the open integral for each problem. Once completed, calculate the required area between the limits numerically. Then choose a graphical method for one by parts and one substitution method. Find the value by graphical means, then compare answers for accuracy with calculated value."

The first question is cos(5x+3)dx in the limits of 0.8 and 0.4. I calculated my answer using substitution to get 0.323... but I am unsure what the question means by "Solve Graphically". I've googled but no real answers that would fit my level of work. Do they mean using squares as area to estimate, or is there a specific method I'm missing?

[lean sun]

In the previous question, it asked me to solve some integrals numerically. I used simpsons/trapezoid/midpoint rule for these three. Does graphically mean finding area by using space on a graph of my curve?

<@&286206848099549185>

[slim jungle]

maybe they are asking you to compute the integral by sketching the graph of the function?

but then im bot sure how are you supposed to find are under a cosine function

[lean sun]

thats what I was confused about

I plotted it, but it's not going to get me an answer

[slim jungle]

we did some integrals by integration and by finding area under a curve using geometry

but integrands were not cosine or any trig functions

[lean sun]

I mean I could use the squares and do a rough estimate but im not sure if a particular method is being asked for

[slim jungle]

they were like a line segment or some kind of shape that u can find area of using geometry(triangle, trapezoid, etc)

[slim jungle]

why take finite rectangles and estimate when u can just integrate and have exact answer

[lean sun]

They want both for some reason

and a comparison

it's so weird

it's this engineering course i'm taking before university

[slim jungle]

really im also confused by what they want you to do, sorry for not helping

[lean sun]

I'd be able to ask my tutor but he's not in today so I'm getting the work done at home

Yeah, it's really confusing

[slim jungle]

is there a chance of asking your teacher maybe?

maybe he can clarify

[lean sun]

I've sent an email but it's a random chance if he replies hahaha

[slim jungle]

or maybe you did this kind of stuff in class

[lean sun]

We used numerical methods in class

I'm wondering if he just miswrote on the page

because using a numerical method like the simpson rule makes a lot more sense than trying to calculate it graphically

especially for such a simple integral

I might just provide a graph alongside the solution using a numerical method

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[ashen thorn]

How do you prove that for any real number x, x^2 > 0 if x isnt 0? (in the simplest way possible)

[mortal trellis]

depends on what you have access to

if x is positive then x^2 > 0 is hopefully clear. if x is negative, then -x is positive and it would suffice to show that (-1)^2=1 is positive

[ashen thorn]

I mean even x^2 >= 0 (for any real x) is hopefully clear to me, I am not sure how to write it down in mathematical language though, I dont quite have access to any fancy tooling, but my teacher wants us to try our best. (it was until now just something we assumed was always true)

[mortal trellis]

x^2=0 means that x*x=0, aka the product of things is zero. that means one of the things has to be zero

unless you want to completely reinvent everything, you have to take certain things for granted

[ashen thorn]

.close

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[zinc haven]

$\int_0^{\frac9{16}} \frac{\dd x}{\sqrt{x+1}+\sqrt{x}+1}$

[zinc haven]

not sure how to approach this

[thorn tendon]

im assuming u typed that out wrong?

[zinc haven]

no

not wrong

[thorn tendon]

ive never seen it as dx as a numerator

but ig it works

[zinc haven]

its not that unusual of a notation

[hollow warren]

yes its quite common

[thorn tendon]

fair enough

maybe its a country thing

idk

[hollow warren]

x = tan^2(theta)

[hollow warren]

^

[thorn tendon]

$\int_0^{\frac9{16}} \frac{\1}{\sqrt{x+1}+\sqrt{x}+1}$ dx

$\int_0^{\frac9{16}} \frac{\d 1}{\sqrt{x+1}+\sqrt{x}+1}$ dx

[ocean seal]

Relu

[thorn tendon]

idk why there is a dot

but thats how ive always seen it

[zinc haven]

because you put \d before it

[thorn tendon]

oh right

$\int_0^{\frac9{16}} \frac{1}{\sqrt{x+1}+\sqrt{x}+1}$ dx

[ocean seal]

Relu

[zinc haven]

but can we please stop about the syntax issues and focus on the problem on hand

[hollow warren]

yeh this is atough one

trig sub gets me nowhere

[zinc haven]

you sure?

[hollow warren]

it doesn't reduce to anything i can solve

[zinc haven]

hm

[hollow warren]

x = u^2

try this

[zinc haven]

so u = sqrt x

[hollow warren]

yes. i do it in terms of x cus its easier to sub in differential

[zinc haven]

$\int_0^{\frac34}\frac{2u}{\sqrt{u^2+1}+u+1}\dd u$

[hollow warren]

multiply top and bottom by the conjugate

[hot hatch]

Hey

[hot hatch]

Multiply by root x+1- (X+1)

[zinc haven]

oh that cancelled out, pretty nice

$\int_0^{\frac34} (u+1-\sqrt{u^2+1})\dd u$

[hollow warren]

erm

denominator should be 2u i think

[hot hatch]

Cancelled with numerator

[hollow warren]

ohh right im blind

here i was solving a different integral

ok should be straight forward from here

[zinc haven]

oh the third term is a bit uhh

problematic

i would have to use hyperbolic trig of this

[hollow warren]

u = tan(theta)

or hyperbolic sub

[hot hatch]

Or u^2+1=t^2

[zinc haven]

hm

udu = tdt

[hollow warren]

trig sub is just easier

ooh maybe not

hyperbolic sub wins this one

[zinc haven]

ok think i got it now

thanks

.close

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[toxic flax]

where is the y^n-1 coming from. integral of y^n dy is not y^n-1

[toxic flax]

oh

im thinking wrong

its undoing the '

s

.close

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[tawdry crystal]

\begin{cases}
3x + 3y + 2z = 3 \
9x + 5y + 8z = 7 \
3x + y + 3z = 2
\end{cases}

A = \begin{pmatrix}
3 & 3 & 2 \
9 & 5 & 8 \
3 & 1 & 3 \
\end{pmatrix}

A' = \begin{pmatrix}
9 & 9 & 6 \
0 & 4 & -2 \
0 & 0 & 0 \
\end{pmatrix}

C' = \begin{pmatrix}
9 & 9 & 6 & 9 \
0 & 4 & -2 & 2 \
0 & 0 & 0 & 0 \
\end{pmatrix}

r(A) = r(C) = 2
\det A = 0

\begin{cases}
9x + 9y + 6z = 9 \
4y - 2z = 2 \
\end{cases}

\text{I choose to use } x \text{ as a free unknown therefore:}

\begin{cases}
x = x \
y = -\frac{2}{3} + 1 \
z = \frac{1}{3}
\end{cases}

[ocean seal]

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

[tawdry crystal]

can someone check this pls

[earnest lark]

The error is coming from each section not being in a maths environment

So you can either enclose each cases/pmatrix with $ signs or use \begin{equation}

Nvm

Am stupid xD thought the error was with complation, I've had a long day ...

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@tender python Has your question been resolved?

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@tender python Has your question been resolved?

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[unborn plover]

can you elaborate

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[patent swallow]

Hi. I need to check if 2 divides 63a'-7b'²

but tbh idk how to continue

oh btw

a' and b' are co-primes

(a' : b') = 1

[rustic coral]

what about a'=2 and b'=3 tho

,w 632-73^2 mod 2

[ocean seal]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[rustic coral]

did you forget a condition like they're both odd

[lone heart]

@patent swallow Has your question been resolved?

[patent swallow]

well they are both coprimes so I guess they cannot be both even

[rustic coral]

coprime just means they share no positive common divisor other than 1

2 and 3 are coprime

(2:3)=1

[lone heart]

@patent swallow Has your question been resolved?

[patent swallow]

so idk

So I need to check if 2 divides 63a'-7b'². I know (a' : b') = 1

This means a' and b' cannot be even at the same time.

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[timber raft]

I found the coordinates of intersection, but I don’t know what to do after pwards

[placid zinc]

What order are you integrating? dxdy or dydx?

[timber raft]

well I have the option of integrating in polar coordinates as well

besides, what would I integrate exactly?

[placid zinc]

We can also do polar! If you know off-hand how to describe that offset circle

[timber raft]

yes but it’s not a circle

[placid zinc]

(x-1)² + y² = 1 is what I was referring to

[timber raft]

Yes

well the area itself would be pi… the radius is 1

but that’s not the question

[placid zinc]

But personally, describing that circle in polar seems weird to me. Rectangular feels more natural

[timber raft]

Sorry I don’t understand what you are trying to say

Do you know how to solve it?

[lone heart]

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[timber raft]

.close

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[cinder elbow]

Somebody please tell me how to solve this

[cinder elbow]

I have been struggling for the past 25 minutes

[brave solar]

take logarithm on both sides

[cinder elbow]

I did

And then what

I already tried that

[brave solar]

solve for x

[cinder elbow]

There are two x so I don’t know what to do

[brave solar]

?

it's a linear equation

you are solving ax+b=0

[cinder elbow]

Wait one second

What do I do after I get: (3x)log6=(2x-3)log2

[brave solar]

it's a linear equation

dont get distracted by those logs

how would you solve a(3x)=b(2x-3)

[cinder elbow]

I would probably divide by 3x on both sides I guess

I don’t know

[brave solar]

collect the x's

[cinder elbow]

What do you mean?

[brave solar]

bring the x's to one side

[cinder elbow]

Like a=b(6x^2-9x)

?

[brave solar]

where's the exponent coming from

[cinder elbow]

Wait I meant to divide not multiply

I messed up

Wait

So a=b(2x-3)/(3x)

@brave solar tell me pls

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[dusky palm]

just wanted to check my work

[dusky palm]

wait just raelized i did it wrong

damn

wait no

idk actually

changed my answer to 2^x-9/4 = y

[hot hatch]

Hey

[dusky palm]

hey

[hot hatch]

Yeah it should be 2 power

[dusky palm]

?

[hot hatch]

$y=2^\frac{X-9}{2}$

[ocean seal]

Monarch of Eternal Night

[dusky palm]

why is it 2 under the x-9 and not 4?

[hot hatch]

@dusky palm

[dusky palm]

yeah i get that part but arent you supposed to divide x-9 by 4 and then the end result would be 2^ x-9/4

[hot hatch]

Oh

[hot hatch]

Yes

That's crct

Mb

I took it as 2

[dusky palm]

oh cool aight

ty

.close

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[dusky palm]

how does one proceed from here

[hot hatch]

Write everything in terms of a^b

For first value

It's $3^\frac{3}{4}$

[ocean seal]

Monarch of Eternal Night

[hot hatch]

Do like this to both numerator and denominator

[dusky palm]

uuuuh i believe i already did that

but idk how to combine them

[hot hatch]

@dusky palm

Use second rule

[dusky palm]

they have different bases though

[hot hatch]

Also denominator is (3x)^1/2

[hot hatch]

Seperate the denominator

As 3^1/2.x^1/2

Now there is 3^m and 3^n in the fraction

[dusky palm]

is the little period between what you wrote meant to be multiplication?

[hot hatch]

As well as x^m and x^n

[hot hatch]

Ah yes

[dusky palm]

gotchu

i got 3^1/4 * x^1/6

can these be simplified further

[hot hatch]

No

That's it

[dusky palm]

okay ty

.close

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[balmy grail]

Divide each number in the row by the total of the row

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[soft finch]

How do I turn off notifications from one of the channels? The help notifications got turned on by default and they're too much.

[alpine sable]

remove the helpers role u have on

can someone calculate the median for this cumulative table Value |Frequency
10, 5
15, 12
20, 20
25, 28

go to "roles and channels" on the top of ur channel list

and opt out of the helpers role

[twin nimbus]

!occupied

[lone heart]

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).

[soft finch]

.close

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[soft finch]

.reopen

[lone heart]

✅

[soft finch]

How do I opt out of a role?

[charred jewel]

#info

[soft finch]

Oh, okay. thanks

.close

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[graceful star]

Is there a faster way I can approach this problem? Or is it just product rule for part A and setting each component equal to zero

[graceful star]

Then Hessian matrix purgatory

[placid zinc]

The actual derivative would be easier as a log derivative

[graceful star]

Hm how would you take that? I've never been taught it

i think its university policy or something to not release answers (but release the questions 💀 ) so im just stuck baffled on whether or not they want me to actually do a million lines of work in 10 minutes

[placid zinc]

It's not complicated. You're just taking the log of both sides before differentiating, and using log properties to make the right side easier to differentiate

[graceful star]

oh i see

in that case it simplifies the algebra? But you're still taking product rule and everything

[placid zinc]

No more product rule

It's a sum now

Keep in mind the chain rule on the left.

The left is log(f)

The partial in terms of x is:
1/f • df/dx

[graceful star]

ln(f(x,y)) = ln(x^2-4y^2)*(e^(-1x^2-y^2))?

then differentiate that

correct

[placid zinc]

Don't forget the log properties to make life easier

Right becomes:
ln(x² - 4y²) + 1 - x² - y²

[graceful star]

ohh okay

i see

thank you 😁

[placid zinc]

We're still in hessian hell though. I don't know if there's a great way to skip it.

At least every term has an f in it now, which can be pulled out of the determinant

[graceful star]

hessian purgatory

https://tenor.com/view/взгляд-2000-ярдов-война-war-soldier-gif-3632617944134077161

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[molten flume]

help

[molten flume]

I don't understand anything about ratical funtions or how to find the Asymtotes, Domain or Range or how

I need to understand how to do this by tomorow morning

[remote heron]

so you have a problem: we can't divide by 0

[molten flume]

this is her work, how she wants it solved

I didn't do any of this, all my teacher

[remote heron]

its really hard to read

[molten flume]

I can try to get a better Pic wat

wait

[remote heron]

but i can see the problem

$y = \frac{x^2}{x^2-4}$ right

[ocean seal]

jan Niku

[molten flume]

yeah

I want to know how to solve each one

[remote heron]

okay

so the problem is you cannot divide by 0

this is what creates the asymptotes

so the task is to find where the denominator (bottom of the fraction) is 0

https://www.desmos.com/calculator/ql0l52hql3

heres a more simple example

the bottom of the fraction is 0 where x=a

so as you move the slider to different values of a, it moves the asymptote

make sense up to there?

[remote heron]

sure

this is a better picture

[molten flume]

no

[remote heron]

no 😭 which part

[molten flume]

yeah it is, no I don't understand

[remote heron]

okay lets start with the easiest example

$\frac 1x = y$

[ocean seal]

jan Niku

[remote heron]

do you know where this has a vertical asymptote?

where is the denominator equal to 0?

[molten flume]

no clue

[remote heron]

for what value of x is x=0?

[molten flume]

okay okay yeah

[remote heron]

hmm?

we can build stuff up from here

but i need your input

[molten flume]

yeah ik ik just one sec I needed to see my notes

[remote heron]

oh, sure, no problem

[molten flume]

Va is the bottom equal to 0 then solve for x right

[remote heron]

it is, yea

[molten flume]

then Ha is determined weather the degree of the numerator is larger or smaller then the degree of the denominater

[remote heron]

yea, HA is a lot more complicated

[molten flume]

so for y= x^2/(x^2-4) how come the VA is X= ± 2?

[molten flume]

okay explain that to me

[remote heron]

lets do the VA first

[molten flume]

yeah alright

[remote heron]

you said it here, we can create the sub-problem

the bottom is $x^2-4=0$

[ocean seal]

jan Niku

[molten flume]

alright

[remote heron]

how's your factoring?

[molten flume]

add 4

[remote heron]

sure

$x^2=4$

[molten flume]

ita decent

[ocean seal]

jan Niku

[remote heron]

this is a difference of squares, if its helpful to know that, but you can also just solve it directly like you are doing

[molten flume]

okay yeah

[remote heron]

we'd take the square root right

it makes a plus or minus

[molten flume]

yeah yeah

[remote heron]

$x=\pm 2$

[ocean seal]

jan Niku

[molten flume]

makes sense yeah okay thanks

[remote heron]

alright

heres the cheat sheet for HA

[molten flume]

alright

[remote heron]

so, whats the case for $y = \frac{x^2}{x^2-4}$

[ocean seal]

jan Niku

[remote heron]

whats n and m?

[molten flume]

they're equal so 1/1 which is 1 right

[remote heron]

so n=m=2

and were in the middle case, I think is what you mean?

and the coefficients are 1/1, so the asymptote is at y=1

[molten flume]

yeah

[remote heron]

okay, any questions on that?

these ... I guess we did two steps?

[molten flume]

yeah

how do I do the domain and range

[remote heron]

domain is easy

range you do some work

this is where that table your teacher makes is pretty handy

do you know what domain and range mean?

[molten flume]

not really no

[remote heron]

so, domain is all of the x values that are 'legal' to put into the function

like for $y = \frac 1x$, the domain is everything except $x=0$

[ocean seal]

jan Niku

[remote heron]

because 1/0 is illegal

[molten flume]

okay makes sense

[remote heron]

and range is all of the y values that are possible to get out of the function

likewise, for $y=\frac 1x$, its everything except 0

[ocean seal]

jan Niku

[remote heron]

because $0 = \frac 1x$ has no solution (it reduces to $0=1$)

[ocean seal]

jan Niku

[remote heron]

wonder how i can draw this

lemme make a desmos thing

[molten flume]

alright alright take your time

[remote heron]

https://www.desmos.com/calculator/qqg1le2odl

the logic is that a function cant cross an asymptote

so, if we test some values around asymptotes, we know what the general shape of the function is

maybe we can take a step back, do you see why we dont have to draw anything to know what the domain is?

[remote heron]

think back to here, these values make us have division by zero

[molten flume]

yeah

[remote heron]

so, the domain is just everything, in the case of the rational functions, except places where we have these asymptotes

[molten flume]

okay that makes sense

[remote heron]

you will get better at guessing values for this the more you practice

in general, you want to try midpoints

like, there are two asymptotes, x=2 and x=-2

[molten flume]

okay

[remote heron]

err, wait, im dumb

[molten flume]

take your time dw

[remote heron]

sorry to be honest these do kind of suck, because its hard to know the middle

but your teacher will probably give you nice functions

[molten flume]

I'm hoping !

I can use a calculator for the points

I can use a calculator for all of it, I'm just not sure how

[remote heron]

actually you know even this isnt true except for vertical

functions approach horizontal asymptotes and avoid vertical ones

[molten flume]

okay makes sense

[remote heron]

wish i had some kind of restricted subset of functions

[molten flume]

I've been trying to find YouTube videos on how to find the points with a TI-83

[remote heron]

at least the two you posted are symmetric

so thats nice

oh man some day i gotta get mine running again so i can help people

i wonder if i have batteries

but even then i have a ti 84

[remote heron]

you mean like maxes and stuff right

[molten flume]

I think so?

[remote heron]

have you tried chat gpt? I mean, to explain to you

dont use chat gpt for math

[molten flume]

the x y chart thay you see there

[remote heron]

but it might be able to help you with your calculator

[molten flume]

yeah real thank you

[remote heron]

im sorry i dont have one in front of me

or id walk you through it

do you get how okay

lets say you could make the graph

[molten flume]

do you know how to do it on a ti-84?

[remote heron]

and you could find the tops and stuff

yea, you got a second?

i think i got some batteries floating around

i just gotta grab them

[molten flume]

aren't they similar with graphing?

take your time yeah

[remote heron]

1s

guess what i found

[remote heron]

okay, you got your calculator?

[molten flume]

yup

[remote heron]

can you just graph the function regular?

[molten flume]

I'll be honest I've never used a calculator to graph

[remote heron]

alright

in the upper right is a button that says "Y="

its right near the screen

[molten flume]

yeah

[remote heron]

you wanna make it look like this

,rotate

[remote heron]

(i typed in your function, note the parentheses)

you know how to do that? you need to use the alpha key i think

[molten flume]

yeah I do I got it

[remote heron]

okay, now hit graph, its on the other side as Y=

if youre lucky, you will see the graph, if youre not, youll see maybe nothing

[molten flume]

I see the graph yeah

[remote heron]

cool

so, it has some cool functions

one is, if you can see the "CALC" thing above trace

if you hit second then that calc

you can find the minimum, maximum, intercept, etc

[molten flume]

alright yeah

[remote heron]

like, say you wanted to find the top of that piece in the middle

you just pick a point to the left

a point to the right

and a guess

you see how this helps you with the domain?

sorry, the range

[molten flume]

yeah I do

[remote heron]

cool :D

you also might wanna note

if you hit 2nd and then the graph button

it gives you a table

[molten flume]

ah thank you

[remote heron]

you can change the jump between values if you hit second then TBLSET (window)

thats about all i can think of to know

[molten flume]

what does the jump between values mean

[remote heron]

so like, if you go to table

theres a 'skip' between all the x values

[molten flume]

yeah

[remote heron]

it might be set to 1 or whatever now

but you might wanna make a smaller or bigger jump

say you want to change inputs by 0.1 every time instead

its called "$\Delta \text{Tbl}$"

[ocean seal]

jan Niku

[remote heron]

this jump

[molten flume]

ah okay

I have 2 questions still and one may seem repetitive but we've been here awhile so you can tell me when you want to stop since I appreciate all your time spent already

[remote heron]

oh, well, whatd you wanna try

i gotta go to bed soonish but I could help you w another

[molten flume]

how is the VA none for the first question?

and I'm still alittle lost on finding the range

[remote heron]

VA means solving where bottom = 0 right

so where is $x^2+4=0$

[ocean seal]

jan Niku

[remote heron]

we can try your thing

$x^2 = -4$

[ocean seal]

jan Niku

[remote heron]

$x = \pm \sqrt{-4}$

see the problem?

[ocean seal]

jan Niku

[molten flume]

yeah that makes sense

[remote heron]

no real solutions

meaning, bottom = 0 doesnt happen

meaning, no VA

[molten flume]

ah okay

[remote heron]

whats the domain

[molten flume]

thank you!

wdym?

[remote heron]

once you know the VA, you know the domain

[molten flume]

x ≠ 0 right or wouldn't be have no domain as well

[remote heron]

the domain is all of the x values that are legal to put into the function

the places where we get illegal values is from VA

(they make division by 0 -- illegal)

but, we dont have any VA

so no illegal points

so, whats the domain?

[molten flume]

no domain right?

[remote heron]

well, its everything

[molten flume]

ah okay

[remote heron]

remember like, 1/x

[molten flume]

so how would i write that?

yeah

[remote heron]

it was everything ,,, except 0

because 1/0 is illegal

but nothing is illegal here

so, its just: everything

[molten flume]

yeah makes sense

[remote heron]

depends on the class

usually something like $(-\infty, \infty)$

[ocean seal]

jan Niku

[remote heron]

have you seen that before?

[molten flume]

ah okay okay

makes sense

[remote heron]

so where are you at

[molten flume]

im all good now!

[remote heron]

https://tenor.com/view/cat-waiting-cat-wait-cat-patient-cat-cute-cat-gif-2663442571218967623

[molten flume]

thank you alot, I appreciate it alot

[remote heron]

no problem

good luck

[molten flume]

thank you!

[lone heart]

@molten flume Has your question been resolved?

[lone heart]

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[lyric mesa]

$4^{2x} = 5^{2x-1}$

[ocean seal]

rynite

[lyric mesa]

how do I solve this? (with exact values)

I found the answer, but as a decimal.

I'm not sure how to find as an exact value.

[elder forge]

4^2x = 5^2x/5

(5/4)^2x = 5

[lyric mesa]

what did you do here

[remote panther]

Stop giving out the answers right away

Explain him instead

[elder forge]

i did not solve it?

[elder forge]

a^m+n = a^m*a^n

[terse stump]

the "-1" in the exponent is taken away by dividing 5^2x by 5
ex: 2^(x-1) = (2^x) / 2
test with any 2

[remote panther]

Do you know power laws @lyric mesa? $a^{m+n}= a^m \cdot a^n$

[ocean seal]

Zue

[lyric mesa]

yes

[elder forge]

i applied that

[lyric mesa]

but 2x-1 are subtracted not added

[elder forge]

just think

it works for both

[remote panther]

Then n would be -1. So you have for that one: $a^{-1}=\frac{1}{a^1}$

[ocean seal]

Zue

[remote panther]

Alternatively one you also adjust the rule to:
$a^{m-n}=\frac{a^m}{a^n}$

[ocean seal]

Zue

[lyric mesa]

I have this so far
$4^{2x} = 5^{2x} \cdot 5^{-1}$

[ocean seal]

rynite

[remote panther]

Yeah good !

No we can bring the terms with the x to one side

[lyric mesa]

$4^{2x} - 5^{2x} = 5^{-1}$

[ocean seal]

rynite

[lyric mesa]

I guess we can factor the LHS

or no...

yeah I can't

[remote panther]

Oh in order to bring the 5^2x over you need to divide

[lyric mesa]

ah I thought it was added

I'll fix

$\frac{4^{2x}}{5^{2x}} = 5^{-1}$

[ocean seal]

rynite

[lyric mesa]

how to simplify this further?

@remote panther

[north perch]

take log on both sides

[lyric mesa]

$\log4^{2x} - \log5^{2x} = \log5^{-1}$

[ocean seal]

rynite

[lyric mesa]

like that?

[north perch]

the base of log is 4/5

[lyric mesa]

what do you mean?

$\log(\frac{4^{2x}}{5^{2x}}) = \log5^{-1}$

[ocean seal]

rynite

[lyric mesa]

I used a log law to rewrite the LHS

[elder forge]

u know that logx^a is alogx?

apply tht

[lyric mesa]

$\log(\frac{4^{2x}}{5^{2x}}) = -\log5$

[ocean seal]

rynite

[lyric mesa]

assuming you were talking about the right hand side

[elder forge]

and left hand side too

[lyric mesa]

how does that apply to the left

[elder forge]

(4/5)^2x

[lyric mesa]

oh I see

[lyric mesa]

$2x\log(\frac{4}{5}) = -\log5$

[ocean seal]

rynite

[elder forge]

yea now solve it

for x

[lyric mesa]

$x = (\frac{-\log5}{\log\frac{4}{5}}) \cdot \frac{1}{2}$

[ocean seal]

rynite

[elder forge]

yes

[lyric mesa]

do I leave the answer like this

$x = (\frac{-\log5}{2\log\frac{4}{5}})$

[elder forge]

yea u can

[lyric mesa]

maybe multiply the 2

[ocean seal]

rynite

[lyric mesa]

alright

[elder forge]

but that is the ans

[lyric mesa]

alr

thank you sm

closing

.close

[lone heart]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[jovial rose]

Is this function continuous?

[topaz lake]

yes

[hushed locust]

consider:

1. where could it be discontinuous?
2. how could you verify whether it's continuous at those points?

[jovial rose]

1. at x = 0
2. I can't ;-;

like, it seems like it's continuous, cuz the limits approach 0

but these are trig functions, and idk a lot about them

[hushed locust]

we say a function is continuous at a point if it satisfies 3 key properties:

1. the function exists
2. the limit exists (this means that both sided limits exist and approach the same point)
3. the function equals the limit

note that sin(x) and tan(x) are both continuous around x = 0, so you can find the sided limits just by plugging in

[jovial rose]

oh okay thank you

I thought it'd be different for trig functions

so it should be continuous at every pi interval right

if I plug in sin(x) < kpi and tanx >= kpi

[hushed locust]

the main thing you have to be careful with for trig functions is at places where they get divide by zero issues (e.g. tan(x) at places where cos(x) = 0). but they are continuous everywhere else

[jovial rose]

alright noted, thank you

.close

[lone heart]

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[lone heart]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[alpine sable]

how do i know AC1 is equal to a/2, all those squares have equal lengths and one square side is a

[lone heart]

@alpine sable Has your question been resolved?

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[ebon sparrow]

!status

[lone heart]

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

[cinder goblet]

1

[ebon sparrow]

have you tried with l'hospital rule?

[cinder goblet]

I'm not allowed to use that

??

[ebon sparrow]

honestly, I was thinking to do something like let u = x^1/3 but then realized it doesn't work lol

it doesn't exist

[cinder goblet]

the asnwer is -1

[ebon sparrow]

thats from the left

but the right isn't

[cinder goblet]

ahh ok

[ebon sparrow]

actually the right should be -1 but the left doesn't have because it doesn't exist

when you graph it you'll see

if $\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+}f(x)$, which means doesn't exist

[ocean seal]

K²en [𝔸𝔨𝔦𝔯𝔞]