help-13

[dusk finch]

if in order for g(x) to be defined, x must be >= 4, then in order for g(h(x)) to be defined, h(x) must be >= 4. So you can actually use a result that you already found.

Or you can just find explicit formula for g(h(x)) and then find the domain as @crimson sedge suggested

[snow hare]

Okay thanks

Wouldn’t it be (2x-8) -4

Since you replace the x

[dusk finch]

yes

simplify that and find the domain

[snow hare]

-6?

[upper ruin]

Pay attention to the sign

[snow hare]

What do you mean?

[upper ruin]

This is 2x - 12, right?

[snow hare]

Yes

[upper ruin]

And you need it to be ≥ 0

So 2x - 12 ≥ 0 => 2x ≥ 12 => x ≥ 6

[snow hare]

So it would have to be positive 6

Thank you

[upper ruin]

Yep

[upper ruin]

Yw

[snow hare]

I can solve this the same way right?

Because g of h is g(h(10))

[dusk finch]

ye, just plug 10 in the formula sqrt(2x-12)

[snow hare]

So sqrt(20-12) which would give me 8

And sqrt(8) is 2sqrt2

[dusk finch]

yes

[snow hare]

Now what is this, I don’t recall seeing one of these?

[upper ruin]

(f o g)(x) is the same as f(g(x))

[snow hare]

f(g(0)) now what do I do?

[upper ruin]

The same as the previous question

Find g(0) and then plug the number you get into f

[snow hare]

Would it be 2 for g?

And -1 for f?

[upper ruin]

g(0) is 2, yep

[snow hare]

Is f right as well?

-1(2(0))

Would this be correct?

[cedar kiln]

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[finite raven]

https://math.stackexchange.com/questions/746147/tangent-line-to-a-differentiable-curve

[finite raven]

Can someone give me a tip

On how to start

I tried using continuity, but I couldn't get anything useful

[cedar kiln]

@finite raven Has your question been resolved?

[finite raven]

<@&286206848099549185>

[finite raven]

<@&286206848099549185>

[finite raven]

<@&286206848099549185>

[thick cipher]

what exactly are you trying to solve?

[wispy lotus]

@finite raven

[finite raven]

I'm trying to prove the result

[thick cipher]

what definition of the tangeant are you using?

[finite raven]

The one in the question

Wait, he doesnt give it

sorry

$L = \left{ \alpha(a) + t\cdot \alpha'(a) : ;: t\in \mathbb{R}\right}$

[wraith dagger]

tales

[thick cipher]

ok, let me think it through

Do you have trouble for the two directions of proof?

[finite raven]

I haven't tried the (<=) yet only (=>)

the definition of d(a(t), L) is wrong as well

d(a(t), L) = inf |a(t) - v|, where v is in L

[thick cipher]

Yes I noticed that it's l in L not alpha(a) in L

[finite raven]

We have showed that is X is closed, there is a x in X such that d(y, X) = |y-x|

So, for every t, there is a point v in L such that d(f(t) , L) = |f(t) - v|

[thick cipher]

Yes, that's true but I don't think that helps because you don't know the behavior of alpha except for it's differentiability

but you can parametrize v by $t' \in \mathbb{R}$ with $v = \alpha(a) + t'\alpha'(a)$. Hence you're looking for $\inf \ |\alpha(t)-\alpha(a) - t'\alpha'(a)| = \inf \ |t-a||\frac{\alpha(t)-\alpha(a)}{t-a} - \frac{t'}{t-a}\alpha'(a)|$

[wraith dagger]

cjg#1618

[thick cipher]

In general you can proceed by inequalities in the quantity you want to find the limit of and you'll be bounded for all $t'$ by $|1-t'\frac{\alpha'(a)}{|\alpha(t)-\alpha(a)|}|$

And by chosing $t' := t-a$ you may figure something out

[wraith dagger]

cjg#1618

[finite raven]

Wait

[thick cipher]

And it might even help you for the reciprocal if I'm not wrong

(I forgot some norms in there but it should still work)

[wraith dagger]

cjg#1618

[finite raven]

I can't make this choice

t' is given, no?

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[deft spear]

would this work?

[dawn junco]

i don't understand what you're going for here

first off, you don't assume the base case, you actually have to show it holds yourself

2. why do you start with n=4

if you want to show that formula works for every natural number n, you'll be missing a few cases

like start with n=1 at least

or n=0 if you feel extremely lazy

[deft spear]

ye i know that was just seeing if a divisibilty proof would work

instead of differentiating

[dawn junco]

well the statement you have to prove involves repeated differentiation

you're gonna have to use derivatives somehow lol

so really the main thing is

assume you know the n-th derivative

how would you compute the (n+1)-th derivative from that ?

[deft spear]

was just thinking if i used a divisibilty proof it might of worked as i got 3f(k). but i know how to do it the other way,

[dawn junco]

aight

so what's that divisibility proof you're thinking of ?

[deft spear]

for example

i tried using f(K+1)- f(k)

for this derivative proof but i dont think it works lol

[dawn junco]

yeah I mean this is a very different question

if the statement involves divisibility, you're prolly gonna use some divisibility rules

if the statement involves derivatives, you're prolly gonna use derivative rules

there's no reason why an approach in one of these proofs would help in the other one really

[deft spear]

alright thak you ill do it the over way lol

other****

[dawn junco]

apart from the fact they're both induction problems, there's not really anything else in common between these two

[deft spear]

thank you

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[distant delta]

How do i solve this

[cerulean star]

tbh, since they gave you four six choices, the fastest way to solve this problem is guess-and-check.

[cerulean star]

1 expand the equation

2 work with it like a polynomial

3 use the Pythagorean Identity

4 rearrange until one side is 2cos(x)sin(x) (which is the double-angle identity for sine) and the other side is constant.

5 use that double angle identity

6 then you will use the unit circle to figure out where 2x is equal to that constant.

7 solve for x

[distant delta]

can i subtitute x for cos and y for sin and work it out and plug it back in?

[cerulean star]

Maybe, at what stage? Show your work

you will be better off using these identities because you are expected to learn and use them

[distant delta]

how do you know to make one side into that double angle identity?

[cerulean star]

because I know it when I see it

when you see cos(x)sin(x), you know you can turn that into double angle identity for sign if you can get a "times 2" in front of it

$\sin{\left(2x\right)} = 2\cos{(x)}\sin{(x)}$

[wraith dagger]

Melvin Eugene Punymier

[cerulean star]

@distant delta how is it going

[distant delta]

im working through it

[cerulean star]

good job!

[distant delta]

what went wrong tho

the answer is different

[cerulean star]

wdym

[distant delta]

^

[cerulean star]

I checked it with Desmos

that is the correct answer

it makes the trig equation equal to 48

LHS

[distant delta]

is -pi/4 a differnt way of writing this in any way?

[distant delta]

how

[cerulean star]

I mean no

1/4 of 2pi is

that would be -pi/2

so that would be equivalent

[distant delta]

so the answer is wrong on the quiz?

[cerulean star]

what does the quiz say the answer is?

[distant delta]

-pi/4

[cerulean star]

actually, that works too

lemme think for a second how we can make that one show itself

oh

I'm just being a dummy

I just got off work

check it out:

3pi/4 + pi/4 = 2pi

✅

that is,

[distant delta]

no that = pi

[cerulean star]

3pi/4 = -pi/4

[distant delta]

doesnt that = pi

[cerulean star]

I drew my picture wrong (pic is fine actually, I just forgot that we are still dealing with 2x for the case of the negative angle as well)

so I am still messing up

remember that we solved for 2x, not x, initially

[distant delta]

yes

[cerulean star]

so we have 2x = 3pi/2 AND
2x = -pi/2

-->

x = 3pi/4 AND
x = -pi/4

do you see?

[distant delta]

how does 2x = -pi/2

oh

[cerulean star]

,rotate

[cerulean star]

You need to draw the unit circle

To visualize it and finish these problems

The positive angle is 3pi/2, so 2x = 3pi/2
The equivalent negative angle is pi/2, so 2x = -pi/2

[distant delta]

how do these end up being the same?

[cerulean star]

rotate the terminal rays back halfway in the other figure, they will point in those directions.

but you shouldn't start with that figure

you are trying to solve sin(2x) = -1

[distant delta]

yes

[cerulean star]

so the angle used in the unit circle picture is 2x, not x

[distant delta]

what should i solve instead?

[cerulean star]

this is where you end up, not where you start puzzling it out

[cerulean star]

👆

your unit circle diagram should be based on that.

[distant delta]

oh

how does that look like?

[cerulean star]

the sine function is ONLY negative on [0,2pi] at 3pi/2.

so the angle 2x = 3pi/2

we can express that as a negative angle instead: 2x = -pi/2

both angles are in the figure

[distant delta]

ok so the answer is both -pi/4 and 3pi/4

[cerulean star]

yes

but the only option they gave you that worked is -pi/4.

[distant delta]

you know like a youtube vid that go through this more

[cerulean star]

that is sufficient

[distant delta]

yea

[cerulean star]

not off the top of my head

[distant delta]

its still fuzzy in my head im a lil slow

[cerulean star]

but this is pretty typical

you treat these long trig equations like polynomials and try to get an equation in terms of one trig expression

you do that by using the Pythagorean and Angle Sum/Difference formulas (usually double angle formulas)

your work is terrific, by the way

very clean, very clear

you just needed to draw the unit circle figure

[distant delta]

thanks

i see it now

to find both the answer on the negative side and the positive side

first I list both the possible sin angles of sin for sin(2x)

then divide both by 2 to solve for x

correct?

[cerulean star]

yes

[distant delta]

ok thanks so much

ive been stuck on this and you helped me get through it

[cerulean star]

there will usually be two angles that satisfy the trig expression you are drawing the unit circle for

HOWEVER

in the case where sin(t) or cos(t) = +-1 or 0, only one angle will make that true (the cardinal angles 0, pi/2, 3pi/2)

but you consider the positive and negative cases anyway when t is an expression (like 2x) and you will get more than one answer anyway

this problem could have had 4 answers (pretty sure)

if the absolute value of the RHS weren't 1...or 0

[distant delta]

like if it was sqrt of x instead of 2x?

[cerulean star]

no

like

if you had

sin(2x) = sqrt(2)/2

[distant delta]

i meant x^2

[cerulean star]

then you would draw a figure where a mystery angle "2x" terminates where sin(2x) is sqrt(2)/2

that happens when 2x = pi/4 and when 2x = 3pi/4

and THEN

you consider positive and negative versions of that angle, since when you solve for x, you will get different answers

so 2x = pi/4 AND 2x = -7pi/4

then x = pi/8 AND x = -7pi/8

then you still need to do 3pi/4 !

2x = 3pi/4 AND 2x = -5pi/4

x = 3pi/8, x = -5pi/8

Four distinct answers

none of them equal to each other

[distant delta]

yk how you do only 1 angle will make it true. But isnt 3pi/2 and -pi/2 same thing? so wouldn't it be 2 angles still?

ok nvm

i get it

thanks for the help!

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[patent oar]

I have made this graph in Desmos that just checks if certain numbers are a part of different groups (like even numbers, prime numbers, etc) and it's been going pretty well, however I had a lot of trouble making an "is equal to" function in which two numbers are inputted and the output is a 1 if they're equal and 0 if not. I think I've come up with something that works, but I'm not quite sure, so could someone just check it please?
Feel free to check the other ones if you want, but I'm pretty sure they're alright.
https://www.desmos.com/calculator/ja4hhxrluw

[patent oar]

<@&286206848099549185>

[cedar kiln]

@patent oar Has your question been resolved?

[patent oar]

Nop

[cedar kiln]

@patent oar Has your question been resolved?

[cedar kiln]

@patent oar Has your question been resolved?

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@patent oar Has your question been resolved?

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@patent oar Has your question been resolved?

[patent oar]

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[snow hare]

I know it’s -1(2(0)) but how do I solve that?

[snow hare]

It says the answer is 3 but how?

[sharp pike]

First find g(0)

then find f at that value

[snow hare]

I did to get -1(2(0))

F is -1 and G is 2

[sharp pike]

You only want to evaluate g at 0

Because $(f \circ g)(0) = f(g(0))$

[wraith dagger]

Ari

[sharp pike]

And you got $g(0)=2$, so $(f \circ g)(0)=f(2)$

[wraith dagger]

Ari

[snow hare]

But how do I get positive 3?

[sharp pike]

what if f(2)?

[snow hare]

3?

.close

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[light grove]

What kind books recommend for calculus learning by my own

[mighty shuttle]

<@&268886789983436800>

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[vocal eagle]

How do i find the equation of a line. I dont know how to find it because one of the point has a missing y coordinate

[smoky idol]

Please give the original problem with all the details

[vocal eagle]

15 is the lenght of the line QS

[livid hound]

you can first apply distance formula to get the y-coord of Q

[ebon juniper]

u can use pythagoream theorem to find the height perhaps

[vocal eagle]

Pythagorean theorem needs 2, but i only have 1 line

[vocal eagle]

How

[livid hound]

do you know the distance formula?

[ebon juniper]

u have base and hypothenuse

[vocal eagle]

Oh yeah

The y coord i got is 10.54

Do i round it up?

[ebon juniper]

i would not round

[vocal eagle]

Okay

[ebon juniper]

idk if 10.54 is right since i didnt go thru the calculations

[vocal eagle]

Yea i doubt its right tbh

[ebon juniper]

thats on u to b right, but anyhow now that u have the y coord

[livid hound]

your value sounds wrong

[ebon juniper]

u can use the formula for a line, y=mx+b

[livid hound]

can you show how you're getting 10.54

[vocal eagle]

=√(15²-12²)

Right?

[livid hound]

no

[vocal eagle]

🥲

[livid hound]

do you know the distance formula?

[vocal eagle]

Nope

[livid hound]

its pretty much pythagoras, do you know that?

[ebon juniper]

this gives u the height of that line on that distance, which is to find the slope.

[livid hound]

draw a horizontal line from Q to ST

[vocal eagle]

Next?

[livid hound]

call the intersection point N

[vocal eagle]

Done

[livid hound]

QN will be 12,
and

=√(15²-12²)
will actually give you SN, which isn't quite the y-coord of Q
but you can actually just determine the slope from here using that
also doesn't explain how you got 10.54

[vocal eagle]

Oh the y is 9

My bad yall

[livid hound]

no

[vocal eagle]

Which one😭

[livid hound]

no

9 is the length of SN

will actually give you SN, which isn't quite the y-coord of Q

[vocal eagle]

Ooooo

So y=12-9?

=3

[livid hound]

yes

[vocal eagle]

Ooo

Thank you both of you

The equation i got is
y = 3/4x + 27/4

How do i close the channel

.close

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[quick flume]

How does that tan^2x derive

[hidden mural]

d(tanx)/dx = sec^2 x

[quick flume]

But on the photo

I don

't get it

[livid hound]

1/cos is sec

1/cos^2(x) is sec^2(x)

they used power chain rule

[quick flume]

What?

ah ok I get it

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[latent birch]

How do I find this

I mean the equation for n

I don't even know where to begin

[lapis flume]

where does the log come from?

[latent birch]

that is a good question

Sorry, there was an error.
there is 1- on the right side of the inequation with >

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[rare lintel]

What am I doing wrong?

[potent echo]

I am here

[wraith dagger]

jandro

[potent echo]

Seems you did correct

@rare lintel

[fresh quarry]

You're correct
The answer is undefined

[rare lintel]

but how do I proceed?

I am given the answer as -infinity

[unkempt kite]

is this the correct answer?

[unkempt kite]

because according to this equation when you put the limits it should be -infinity

as ln(x-2) will become ln0

which tends to -infinity

[rare lintel]

but we don't have any limit, no?

[unkempt kite]

0 to 2 its in the image

[rare lintel]

oh right

thank you

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[snow hare]

Can someone check this answer and let me know if I got it right?

[proper fjord]

the domain of a function is the range of its inverse

[snow hare]

So it would be the first choice?

[proper fjord]

sorry i read the problem wrong, your answer is correct

you can plot it on a graphing calculator to see a visual example proving it's right

[snow hare]

Oh okay I did that and just wanted confirmation

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[quick flume]

At the end in the last 3 lines

from the first to the 2nd where does the -2 go

and from 2nd to 3rd how do you get the final answer in general?

[chrome marsh]

1st to 2nd, a 2sin(x)cos(x) term got factored out

2nd to 3rd uses the double angle formulas sin(2x)=2sin(x)cos(x) and cos(2x)=cos^2(x)-sin^2(x)

[fervent jackal]

@quick flume

[quick flume]

yes?

[fervent jackal]

this

if u got it do .close

[quick flume]

.close

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[quartz crypt]

Not sure what I did wrong on this problem

.close

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[cedar kiln]

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[opaque glacier]

!help

[cedar kiln]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[opaque glacier]

what is wrong with mathway

[dull oxide]

What are you trying to do?

[opaque glacier]

evaluate

but mathway cant solve it

very strange

[dull oxide]

Can you solve it?

[cedar kiln]

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[crimson sedge]

prove that no (x,y,z) exist that verify the following equation :

[austere ember]

there's definitely some x,y,z that satisfies that, i'm assuming you mean in the integers?

[crimson sedge]

ah yes this is in an arithmetics related course

[cedar kiln]

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[lean seal]

Can I have some help with my homework?

[lean seal]

It’s fairly easy, just linear relations

[cedar kiln]

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[wraith dagger]

faiyrose

[mental trail]

if you managed to prove that p = a specific value, then something went wrong

oh wait + 3x

ok nvm

lemme check

(x-3)(x-1/3) = x^2 - 10/3 x + 1

so if p(x-3)(x-1/3) = px^2 + 3x + r

then -10/3p = 3

p = -9/10

checks out

yes

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[cedar kiln]

Please don't occupy multiple help channels.

[whole umbra]

Can someone check my work

[cedar kiln]

@whole umbra Has your question been resolved?

[whole umbra]

<@&286206848099549185>

[cedar kiln]

@whole umbra Has your question been resolved?

[whole umbra]

<@&286206848099549185>

[jade ridge]

Hello

What problem is troubling you?

yes i check

[whole umbra]

Nothing in particular

[whole umbra]

Wait was 3 relative max

[jade ridge]

no i read the question wrong wait

[whole umbra]

My biggest concerns r 5-10

[jade ridge]

it is relative min

[whole umbra]

K

[jade ridge]

ok i look at those questions might take a little while (:

[whole umbra]

Tysm

[rough terrace]

@whole umbra What is the definition of relative maximum?

[jade ridge]

5a and 5b is correct

but 5c

hmm

on graph f'(x)

its correct that f(x) dosent have a maximum

5d is correct

but your infinity sign is har to read thought it was 8 lol

yes all of problem 5 seem fine

[whole umbra]

Uhmm like the highest point in an area

[whole umbra]

Oh yeah I’m bad with those

[whole umbra]

For 6 can u check b

7 graph I’m a little shaky in

And 10

[jade ridge]

sorry no

[whole umbra]

Np

<@&286206848099549185>

<@&286206848099549185>

[crimson sedge]

!15mins

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[fading zodiac]

What do u need help at

[whole umbra]

What

[whole umbra]

I was able to complete it

But I just need someone to check the work

6b

7

And 10

[fading zodiac]

On it

[cedar kiln]

@whole umbra Has your question been resolved?

[whole umbra]

Update?

[charred coral]

Can someone help

Can someone help

[coral jewel]

dude shut it

[cedar kiln]

@whole umbra Has your question been resolved?

[onyx shard]

Um

Do u know the expression of a circile?

like x^2+y^2=r^2?

[onyx shard]

..

[charred coral]

Yes

[onyx shard]

and

Then, do u find this function familiar?

It is a semi circle

[wicked mantle]

!help

[cedar kiln]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[wicked mantle]

Stop raiding other’s channel

[cedar kiln]

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[obsidian mauve]

Am I supposed to convert some to cosine?

[proven turtle]

!help

[cedar kiln]

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[obsidian mauve]

.close

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[earnest geyser]

Yes

Convert it to cos

And get a quadratic equation

@obsidian mauve

[obsidian mauve]

But how can do I it?
since cos(pie/2-angle)

.reopen

[cedar kiln]

✅

[earnest geyser]

sin^2+cos^2=1

So take sin^2(x) as 1-cos^2(x)

[obsidian mauve]

So cos^2(x)-1=0

And then just do the regular trig methods?

[earnest geyser]

Yes

Wait

One min

[earnest geyser]

How did you get this?

[haughty osprey]

This has to be made quadratic with either sin,cos or tan

[obsidian mauve]

sin^2(x)+2cos(x)-2
= 1-cos^2(x)+2cos^2(x)-2
= 1-2+2cos^2(x)-cos^2(x)
= cos^2(x)+1

[haughty osprey]

If substitution doesn't work , try division

[earnest geyser]

The middle one is just 2cos x

[haughty osprey]

This is wrong

[haughty osprey]

Yea

[earnest geyser]

Not 2cos^2(x)

[obsidian mauve]

Ok then

1-cos^2(x)+2cos(x)
= -cos^2(x)+2cos(x)+1
= cos(x) [-cos(x)+2]+1

[earnest geyser]

-1

[obsidian mauve]

mb

[haughty osprey]

It'll be -1 on the second step

Also you have to solve by splitting the middle term or using the quadratic formula

[earnest geyser]

So it'll be. $-1+2\cos x -\cos^2 x=0$

[obsidian mauve]

Gimme a minute

[wraith dagger]

Monarch of Eternal Night

[haughty osprey]

It's pretty easy from here

[earnest geyser]

Yes

[obsidian mauve]

SHould be easy from here if I didn't make silly mistakes

[haughty osprey]

Yea this is correct

[earnest geyser]

So cos x=?

[obsidian mauve]

0 or 360 degrees

And then the second part

It's the same

Got the answer

Thanks a lot man

.close

[cedar kiln]

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[keen crest]

hello, i need some help on 3d trigonometry, i cannot figure out how to find DX in (a)(ii)

[cedar kiln]

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[ancient lodge]

.reopen

[cedar kiln]

✅

[ancient lodge]

Find AX in triangle ADX (from AD) and use that to find DX

[cedar kiln]

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[somber pagoda]

Hey , I need help with Stereometry. Anyone who understands this? I really need help. tomorrows a exam. :)

[cedar kiln]

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@somber pagoda Has your question been resolved?

[somber pagoda]

Please explain in detail and show the procedure for solving the following 4 math problems. 1. The sugar cube has dimensions of 1.2 cm * 1.2 cm * 1.2 cm. How many sugar cubes are needed to make a large sugar cube with a volume of 216 cm x 3? 2. A regular four-sided pyramid and a cube have bases of equal areas. Determine the volume of the pyramid, if the volume of the cube is 343 cm x 3 and the height of the pyramid is twice the height of the cube. 3. Determine the area of ​​the axial cross-section of an upright cylinder with a volume of 81 pi cm at 3 whose height is three times greater than the radius of the base. 4. Determine the area of ​​a regular six-sided prism, if the surface of the base is equal to 723 cm x 2 in length, the height of the prism is twice the length of the base edge. 5. Calculate the area of ​​an upright cone if the radius of the base is r = 15 cm and the mantle P = 375 pi cm at 2.

[cedar kiln]

@somber pagoda Has your question been resolved?

[cedar kiln]

@somber pagoda Has your question been resolved?

[somber pagoda]

?

What?

dont see the image and why did you delete the msgs

[cedar kiln]

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[median jolt]

Why is the -1 needed?

Can't you just do -10x2?

Wouldn't that still be negative?

[astral bay]

well (x-10)(x+2) is x^2 - 8x - 20

so we did get the -20, the issue is that we ended up with -8x

[median jolt]

So it affects the other parts of the problem?

[astral bay]

yeah, the challenge is to find a pair of numbers that gets the right values for b and c

[median jolt]

Oh

[astral bay]

it's very easy to get either one individually, but we need both, because "well it's (x-10)(x+2) + 7x" isn't particularly helpful

[median jolt]

Kk

Tyy

Oh wait

The -1 though

Why specifically -1?

[astral bay]

-x is the same as -1x

if it was something else, like -4x, we'd want the numbers to sum to -4 instead

[median jolt]

Oh

Ok then

Ty

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[cerulean hamlet]

🙏 what x and y values give me -1/rad 3 for tan

[cerulean hamlet]

This is what I have so far but I don’t know how to divide them….

[cedar kiln]

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[cerulean hamlet]

.close

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[crimson sedge]

im kinda confused on what to do after this, or if this is correct

[subtle harbor]

you forgot the +20?

or did you

your work is a bit messy

you have the right idea thought, subbing in x(t),y(t),z(t) of the line in for x,y,z of the plane

I just think you forgot the +20 at the end though

[crimson sedge]

ya its messy, theres a +20

sorry about that 💀

[subtle harbor]

here specifically

[crimson sedge]

OHHHH yeah

i did

okay thank oyu

i am just confused what getting t = 0 implies as a case solution

[subtle harbor]

well

you have a line with t as a parameter

so whadya think you should do

[crimson sedge]

im not sure, i just assume infinite solutions if 0t=0

because of real values

[subtle harbor]

wait how are you getting 0t

i thought you said you got t=0

[crimson sedge]

expanded the left side and got 0t = 0

maybe i should properly express that in my work 💀

[subtle harbor]

well if you get infinitely many solutions, that means there are infinitely many solutions, i.e. intersections between line and plane

which implies...?

[crimson sedge]

every point on the line intersects

[subtle harbor]

so what would the intersection of the line and plane be

(its not a trick question)

[crimson sedge]

the vector equation from earlier

[x,y,z] = [2,4,10] + t[2,3,3]

[subtle harbor]

i.e. the line itself

[crimson sedge]

yass

[subtle harbor]

there ya go

[crimson sedge]

ok thanks for helping out

[cedar kiln]

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[tranquil summit]

A little confused on why the answer is only the positive root and not both the positive and negative root.
It would be much appreciated if the algebraic method is used, and not the right-triangle method.

[humble karma]

arcsin has range [-pi/2, pi/2], so cos(theta) has to be >= 0

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[digital cliff]

okay

[short moth]

any luck?

<@&286206848099549185> Pls help

[gilded laurel]

Is E the answer? @short moth

[short moth]

ya

[gilded laurel]

Just use the values in the function

[short moth]

avg time to solve is 3 mins

it will take too long

• no calculator

[gilded laurel]

You'll get the maximum value from option E

There's a trick btw

[short moth]

hm?

[gilded laurel]

Wait no calculator?

[short moth]

nop

[gilded laurel]

Whoops

Sorry my bad then

[short moth]

nw

[gilded laurel]

Where do you study?

[short moth]

india

but

this is cambridge's exam

[gilded laurel]

So you're preparing for admission? Or scholarship?

[short moth]

admission

[gilded laurel]

Oh r u studying in college?

[short moth]

no no

highschool only

this is the tmua cambridge exam

just critical thinking math

[gilded laurel]

Oh, I understand

You can memorize the values of degrees of cosine and sine functions

That'll help

Like cos 270 is 0

sin -90 is -1

So this will make help you make quick decisions fast

[short moth]

but

it doesnt help here

.x close

.close

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[heavy edge]

what is the value of y

[twin finch]

extrapolate cd

youll get it

[heavy edge]

.close

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[slate garden]

For 6 a

Why does my method not work

Why do I have to sub in x = 8

Instead of doing what I did

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[indigo crag]

Can anybody explain why the number of functions from N to {0, 1} has the same cardinality of the powerset of N?

[rigid lava]

for a fixed n, there's two options for f(n), 0 or 1

[slate lintel]

you can map between the two pretty easily. The powerset of N is the set of all subsets, and if you have some subset of N (call it A), you map that to the function where f(n) = 1 if n is in A, and f(n) = 0 if it is not

[indigo crag]

Oh right

thanks!

to you both

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[void sand]

.close

[solemn idol]

Is it true the following is A relation that is reflexive and symmetric but not transitive -

Let aRb iff a and b are integers that have common dividers other than 1 (or equivalently gcd(a, b) > 1). This is clearly reflexive and symmetric but not transitive right?

[dusk finch]

hmm

is 1R1 true?

this one example shows that it's not reflexive

and yes, its not transitive either

[cedar kiln]

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[wicked mantle]

@solemn idol Are you ok with his explanation?

[cedar kiln]

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[bold pond]

can someone help me with master method?

[cedar kiln]

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[bold pond]

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[mental trail]

divide by this

-5* 6/5 * cbrt(a/2) ^3 = 6-a

notice that cbrt and ^3 cancel out

[cedar kiln]

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✅

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[cedar kiln]

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[sinful bay]

!xy

[cedar kiln]

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

[vernal basin]

Yes

Also because of law of sines

[cedar kiln]

@crimson sedge Has your question been resolved?

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[cedar kiln]

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[viral bobcat]

part a

[upbeat dune]

Angle of depression should be from horizontal

So draw a horizontal from rhe top of the cliff

[viral bobcat]

alright but i think its the angle of the top of the cliff that faces the 100 meters

[supple mountain]

its not because that would be unsolvable prty sure

[cedar kiln]

@viral bobcat Has your question been resolved?

[cedar kiln]

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[thorn wyvern]

Sum1 help me solve this pls

[thorn wyvern]

@pale bronze please carry me

<@&286206848099549185>

He’s abusing me

Fake math help man

.close

[cedar kiln]

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[mighty shuttle]

<@&268886789983436800>

[dusk finch]

Hm?

how did you even know about them?

[meager wagon]

can u make the question more readable please

[thorn wyvern]

He was trying to clown me for getting math help

[thorn wyvern]

I’ll type it out

[meager wagon]

aight

[dusk finch]

The city is bulding a road (gray) that will take away part of your property the road runs perpendicular to side AD and intersects AD at point E(18, -8). The city has agreed to put a fence along EF. What would be the length of the fence?

@thorn wyvern can you just check if the values agree?

[thorn wyvern]

Point E is 18 -8

My friend rewrote the question and mixed it up so it’s a little confusing

I’m a little unsure on how to solve for the distance of FE

[fiery thicket]

you could treat the points as position vectors to then find the vector FE

just looking at it rn

[thorn wyvern]

Wdym position vectors

[dusk finch]

My personal approach would be finding equation of the line perpendicular to AD at point E

then equation of the line BC

and the intersection would then be F

[thorn wyvern]

O so like find the slope of AD?

[dusk finch]

yes

and then slope of the perpendicular line would be -1/m

where m is the slope of AD

[thorn wyvern]

Slope of AD would be -11/15 right?

[dusk finch]

A is (3, -3) D is (21, -9)?

[thorn wyvern]

O wait messed it up

I put e instead of d

-6/18 right

[slim jasper]

.reopen

[cedar kiln]

✅

[dusk finch]

oh thanks

[dusk finch]

mhm

B (11, 21)
C (29, 3)

(please check if i read those coords correctly)

[thorn wyvern]

Yep

[dusk finch]

also E (18, -8)

[thorn wyvern]

Ye

[dusk finch]

cool, so if slope of AD is -6/18

whats the slope of line perpendicular to it

[thorn wyvern]

Thts 18/6 right so it’s 3?

[dusk finch]

correct

so we are looking for an equation of line with slope 3, that passes through E

[thorn wyvern]

So y=3x +b

[dusk finch]

yep

[thorn wyvern]

So do I sub in point e

?

[dusk finch]

yeah, try that

[thorn wyvern]

-62 is b?

[dusk finch]

Yep, so now we have y = 3x - 62

now to find F, we also need the second line

which passes through BC

B (11, 21)
C (29, 3)

[thorn wyvern]

So find perpendicular bisector?

[dusk finch]

hmm?

[thorn wyvern]

Same thing I did w AD?

[dusk finch]

wait no

you dont need perpendicular line anymore

[thorn wyvern]

O

So normal slope?

[dusk finch]

ill draw diagram

wait a moment

[dusk finch]

yep, start with that

note that we currently have equation of the red line

[thorn wyvern]

Slope is -1

[dusk finch]

to find point F, all we need is find the equation of line BC (blue line) and then find the intersection of the 2

[thorn wyvern]

For bc do u sub in one of them to find the equation?

To find the equation

[dusk finch]

yes

[thorn wyvern]

Ok I got y=-x +32

[dusk finch]

yep

so now find the intersection of those 2 lines

[thorn wyvern]

Ok

[dusk finch]

y = -x + 32
y = 3x - 62

[thorn wyvern]

I’m a little lost

I get rid of y first

Then it should be -4x + 94 then -4x = -94

Right?

[dusk finch]

yep

so x = 94/4

so x coordinate of point F is 94/4

[thorn wyvern]

Is it fine for it to be a decimal?

[dusk finch]

yes, its fine

its 47/2 if im not mistaken

which is 23.5

[thorn wyvern]

Ye

So 23.5 and I sub in

[dusk finch]

yep

so y = -23.5 + 32

[thorn wyvern]

Ye so 8.5

[dusk finch]

correct

so now you have coordinates of E and F

[thorn wyvern]

So f is 23.5 and 8.5

[dusk finch]

and you can calculate the distance

its a bit tedious calculation

,w sqrt((18 - 23.5)^2 + (-8 - 8.5)^2)

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by methisalwaysright

[thorn wyvern]

Tht parts easy after tht

Tysm for the help

.close

[cedar kiln]

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[dusk finch]

np

[cedar kiln]

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[ember pumice]

h

[cedar kiln]

Please don't occupy multiple help channels.

[ember pumice]

anyone can help with this?

[sterile vine]

Brother u got ur answer

Use the respective surface area formulas

For each shape

[ember pumice]

he

how though

i just got into this class rlly late

a sphere

[wraith dagger]

faiyrose

[ember pumice]

how do i round them to the nearest tenth

how do i put that in a calculator ?

okay what next

[sterile vine]

Are you familiar with radius

[cedar kiln]

@ember pumice Has your question been resolved?

[cedar kiln]

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[sterile wedge]

this is a question in ODE I need to find f(x), above is what I tired

[mighty shuttle]

let $f(x)=y$

[wraith dagger]

ƒ(Why am. I here)=I don't know

[mighty shuttle]

then we have $2x\frac{dy}{dx}+y=y^2$

[wraith dagger]

ƒ(Why am. I here)=I don't know

[mighty shuttle]

can you solve this?

oh, and find the inital value too

by subbing x=0

[sterile wedge]

we are asking to find a general function no intial value given

[sterile wedge]

I'll try..

however there is a singular solution being y=0 for all x

[mighty shuttle]

wdym

[sterile wedge]

https://en.wikipedia.org/wiki/Singular_solution#:~:text=A singular solution ys,some point on the solution.

[mighty shuttle]

$f(0)=f^2(0)$

[wraith dagger]

ƒ(Why am. I here)=I don't know

[mighty shuttle]

yes

[sterile wedge]

actually now that I look at my answer I can see a mistake

lol I did df/dy XD

next time I should just substitue f(x) =y

I think I can solve it by myself now

thanks

.close

[cedar kiln]

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[mighty shuttle]

👍

[cedar kiln]

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[shut wadi]

none of the triangle congruency criteria I think

so do I put 'not enough info'

yeah

because

the bottom side

we don't know its length

so it can go as far as it wants

and the left side

wait no

hm

nvm

even if the bottom angle went further

the left side

we know its length

so it can't reach

the bottom angle

and become a triangle

hm

it can't be ass

ssa?

does that exist

no

[hollow trail]

side-angle-side is a valid triangle congruence criterion

[shut wadi]

that is what I was thinking

but how

like

wouldn't sas

be like this

[hollow trail]

the two marked sides are shown to be equal in length (congruent), and i believe the marked angles are meant to be congruent as well

[sterile vine]

There’s two sides already

Why need a third side

[shut wadi]

ik

[sterile vine]

sas

[shut wadi]

but shouldn't the 2 sides

surround

the angle

[sterile vine]

Brother ur overthinking this 💀

[shut wadi]

like

[hollow trail]

the angles touch corresponding sides (2 marks and unmarked)