help-36

[dense coral]

<@&268886789983436800>

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[jade fable]

I hace a design of a math app

[jade fable]

Where can I discuss it

Where should I post it

[proper dagger]

#math-discussion maybe

[jade fable]

.clo

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[digital turtle]

umm I don't know how to set up my theta equation

Like for example sin theta= y/z

I know how to set it up when I have dy/dt and dx/dt but not when I only have dz/dt

This is a related rates problem

[final saddle]

@digital turtle Has your question been resolved?

[formal trail]

you have one more derivative that you haven't considered yet

[digital turtle]

dtheta/dt?

[formal trail]

well that's what you're trying to find

[digital turtle]

Uhh

[formal trail]

you have one more derivative which is given to you in the information

[digital turtle]

I'm reading it I can't find anything

[formal trail]

it's not given explicitly, but you can figure it out from context

[digital turtle]

do I use PT?

[formal trail]

a thing to consider is, what quantities are constant in this problem

[digital turtle]

Oh

Is dy/dt

The constant

I mean

It'll be zero

But is y my constant

[formal trail]

yes, exactly

[digital turtle]

ok wait let me see if I can solve it now

Wait

I still can't solve it

Cos(theta) •dtheta/dt=1/z(dy/dt) right?

[formal trail]

not quite

since y is a constant you can just treat it as one

but you are neglecting that z can change with time

[digital turtle]

I do not understand

[formal trail]

when you took the derivative of y/z as 1/z dy/dt that is basically treating z as constant

but z isn't constant

[digital turtle]

Ohhh

Oh my god

I've just been doing that because z has been my constant for like every other angle problem

[digital turtle]

Is this quotient rule?

Or I mean do I use it

[eager mortar]

you shouldn't need it for this problem

[formal trail]

that's not applying the power rule

[digital turtle]

Oh I use power

Wait why

[eager mortar]

where are you getting y/1 ?

[digital turtle]

umm i don't know.. 😅

I was able to solve another similar problem where z was my constant instead of y

Idk why this one's giving me so much trouble

[eager mortar]

i think you need to write out more of your steps, im not seeing your entire process here
cos(theta) dtheta/dt is correct, how did you get there?

[digital turtle]

I derived sintheta

[eager mortar]

okay, why

[digital turtle]

sintheta=y/z

[eager mortar]

so you differentiated both sides with respect to t and got cos(theta) dtheta/dt

but what about the other side

what is the derivative of y/z with respect to t

[digital turtle]

That's what im struggling with

[eager mortar]

okay, well when we say "differentiate with respect to t", its because the expression we are differentiating is actually a function of time

if it wasn't itd be constant with respect to t and the derivative would be 0

so in what way is y/z defined with respect to time?

sin(theta) is defined with respect to time because the angle is changing every second.

[digital turtle]

y/z defined with respect to time because

[eager mortar]

that might help you figure out how to solve d/dt y/z

[digital turtle]

I mean

z changes as like

it's reeled in

[eager mortar]

which is a process that happens over time

[digital turtle]

Yes so in respect to time

But

How does that give me the derivative

[eager mortar]

you know that z is a function of time

so derive

$\frac{\d}{\dt}\frac{y}{z(t)}$

[soft zealot]

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

[eager mortar]

oops

[whole halo]

try not to use \d and \dt like that

what even are those commands for

,,\d test

[soft zealot]

mtt

[whole halo]

,,\dt test

[soft zealot]

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

[whole halo]

,,\frac d{dt}\frac y{z(t)}

[soft zealot]

mtt

[digital turtle]

and y is our content here

Constant

[eager mortar]

havent done latex actively in like 9 months but couldve sworn there was something to make the d upright

[whole halo]

you have to define that command yourself, popular option is \mathrm d or for this server \dd{t}

[eager mortar]

yes, the reason i had you express it like this is because it becomes obvious what your constants are when you explicitly label what is and isnt a function of time

use your derivative rules

[digital turtle]

doesn't this just turn the right side to 0

since 0 is the numerator

[eager mortar]

no

is the derivative of 1/x with respect to x 0?

y is a constant, the quantity y*z(t) is not

[digital turtle]

no its 1

[eager mortar]

no its not

1/x = x^-1 then power rule -x^-2 = -1/x^2

[digital turtle]

wait

can you relate what you're saying

To this

Hold on

Getting you the word problem

[eager mortar]

this is the right half of what you get after differentiating sin(theta)

your current right half of that equation is wrong

if it was right you could move onto the next step

[eager mortar]

dont need the word problem

[eager mortar]

you dont seem to know how to differentiate this

even outside the context of the problem

but that is required knowledge

[digital turtle]

y•z^-1?

[eager mortar]

thats another way of writing the same thing, not the derivative

[digital turtle]

so that's -y•z^-2

-y/z^2

[eager mortar]

yes

[digital turtle]

I don't know how I derived in this problem so easily but why I'm struggling with the problem were on right now so much

[eager mortar]

so back to the original thought process

you relate the sin of the angle to the ratio of the opposite side (y, constant) over the hypotenuse (line length, changing with time)

$sin(\theta) = \frac{y}{z(t)}$

[soft zealot]

Karma

[eager mortar]

then you differentiate with respect to time, because the goal is to figure out how the angle is changing with time

$sin(\theta)\frac{d}{dt}=\frac{y}{z(t)}\frac{d}{dt}$

which gets you what we just worked out

$cos(\theta)\frac{d\theta}{dt}=-\frac{y}{z^2(t)}\frac{dz}{dt}$

[soft zealot]

Karma

[eager mortar]

do you follow so far

this is where you should be at

[digital turtle]

I have that but I only have z^2 not the (t)

[eager mortar]

its the same thing, im just denoting that z is a function of time, but you can just write z

[digital turtle]

Can I plug in at this point

[eager mortar]

not quite

what are we trying to find? we want the rate of change of theta with respect to time, is that anywhere in our equation?

[digital turtle]

This problem is confusing me so much more than the other but it's the same layout

[digital turtle]

Yes

[eager mortar]

which term is it

[eager mortar]

which term in the equation is the derivative of theta wrt time?

[digital turtle]

wrt?

[eager mortar]

with respect to, shorthand. just tired of typing it lol

[digital turtle]

I mean for my other problem I just

Divided both sides by

costheta

And got the answer

I thought I could do that here too

[eager mortar]

yeah but you dont have theta

[digital turtle]

oh well I mean I'd have to plug in my values yeah

[eager mortar]

so how would you plug in values

but you dont have theta

[digital turtle]

cos(20/25)

is it not

[eager mortar]

cos(theta) is adjacent / hypotenuse
the adjacent side is shrinking.

actually

it might still work i might be doing too much

[digital turtle]

I am so lost

[digital turtle]

Should I try it

[eager mortar]

yeah actually it should be fine

[digital turtle]

okay so basically my main slope in this problem was just

Deriving

y/z

[eager mortar]

that and recognizing what your constants were

[digital turtle]

Okay um can you help me understand why the derivative for this problem was -y/z^2,
but for my other problem, it was as easy as 1/z(dy/dt)

[eager mortar]

thats why instead of z i used z(t)

[eager mortar]

i dont really know what your other problem was but the derivative of y/z just is (-y/z^2)

you had to derive y/z

so you got that

simple as

if the problem was different and you had to derive something else, then you wouldnt have had to derive y/z and wouldnt have gotten that

[eager mortar]

because this last problem was of the form

$a * \frac{1}{f(t)}$

while that one gives you something that looks like

$f(t) * \frac{1}{a}$

[soft zealot]

Karma

[eager mortar]

in the first case, the function is in the denominator
in the second, the fraction 1/a is a constant.

the derivative of

$Cf(t)$ where $C$ is a constant (and $f$ is linear) is just $C$

so if you do some algebra for the sake of demonstration:

$C = \frac{1}{a}$

the second example becomes

$Cf(t)$ and so the derivative is $\frac{1}{a}$

[soft zealot]

Karma

[eager mortar]

the point being, the derivative becomes less "clean" when the variable is in the denominator

the reason that doesnt "apply" to the second case is because the variable isnt in the denominator, which means we can represent most of the fraction as a constant

and then the power rule removes the function and leaves you with the constant

[digital turtle]

Wow that makes a lot of sense

Thank you so much

that was a rough journey

Have a good night!

[eager mortar]

lol

u2

[digital turtle]

I feel like that shouldn't have been as difficult as it was for me...

.close

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[delicate trench]

struggling a little bit, how would i convert x+3 to x^2 + 5x + 6?

[final tangle]

consider factorising the x^2+5x+6

or dividing x^2 + 5x + 6 by x+3

[delicate trench]

well its that i have to end with x^2 + 5x + 6 as the denominator

[final tangle]

yes

[delicate trench]

does that work fine if i factor them

ya ig it would

[final tangle]

and doing what i mentioned will indicate
(x+3) * what = x^2 + 5x + 6

[delicate trench]

on it

so (x + 3) * (x + 2)

and numerator adds to x^2 + 7x

and its right boom

ty

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[delicate trench]

.reopen

[final saddle]

✅ Original question: #help-36 message

[delicate trench]

why is this wrong

wouldnt i just be multiplying by x+5

[pliant shore]

how did you get the numerator?

[delicate trench]

wait

idiot brain strikes again i thought we were adding

.close

[pliant shore]

yeah I figured

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[delicate trench]

whats the difference between an extraneous solution and a wrong answer

[delicate trench]

.close

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[vestal crest]

oh.

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[fair sigil]

yo

[fair sigil]

i just need a hint solving the problem

[gritty chasm]

ok post the problem then?

[fair sigil]

yeah i m posting

I need a hint to solve the concurrency point

figure on left

I need a hint to solve the concurrency point

[opal plinth]

Do you have an idea of the position of N?

[fair sigil]

i have drawn it in my nb

[opal plinth]

Not my question

[fair sigil]

<@&286206848099549185>

[proper dagger]

hm, but you haven't answered the previous helper's question

[fair sigil]

i hv drawn it in my nb

[proper dagger]

but the helper said that that's not what they're asking, no?

[fair sigil]

<@&286206848099549185>

[severe verge]

since we know OA'DH is a trapezoid with OA'//DH, if I construct the perpendicular bisector of A'D, which point on OH would that perpendicular bisector go through?

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@fair sigil Has your question been resolved?

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[blazing fable]

Can any helpers help me to solve this problem?

🙁

I’m actually not sure

[cursive bough]

it usually helps to start with smaller cases. consider the number 777 (3 7's). what if you take the square root of that number? it will be about 27.9. figuring out the first digit (in this case 2) isn't very difficult actually.

since you only need the first digit, think about powers of 10. it should be pretty clear that the square root of 777 is between 10 and 100. you just need to test 2x10, 3x10, 4x10, etc, and square those numbers

for example, if you square 20 you get 400, and if you square 30 you get 900. so the square root of 777 has to have the first digit "2"

you can do something similar for cube roots and higher roots

[blazing fable]

Oh

[final saddle]

@blazing fable Has your question been resolved?

[blazing fable]

Just wait a second. Bot

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@blazing fable Has your question been resolved?

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[velvet pivot]

Can I get help with this question about finding the a value in a quadratic function and also with this other question that I don't even know what it's looking for? I tried to substitute -2 and 18 into the factored form of the quadratic function a(x-4)(x+8) by making it 18=a(-2-4)(-2+8) which should equal 18=12a and that should simplify to a=3/2 but the multiple choice had a negative 3/2 so I chose that. And for the 2nd question I got (x+1) as the y value in a coordinate plane and (2x+0.5) as the x thing and if we're supposed to take the formula for a triangle then the base would be halved which would equal 1x+0.25 and when that's multiplied with the height of x+1 we should get answer C?

[velvet pivot]

If you let y represent the north direction of their trajectories and let x represent the west trajectory from the camp, then at x = 1 (because in the question x represents Lindsey's distance) it would be the function of f(1) = 2(1) because 2x represents the distance in 2 pm how far away the other girl is away from the camp compared to the first girl right?

[late rose]

you likely had a calculation error for the first question

everything you did was right

[velvet pivot]

Wouldn't replacing one of the vertices as in like instead of inputting -2 and 18 which is the vertex for something like one of the x intercepts just result in a = 0?

[late rose]

however (-2-4)(-2+8) is not 12

[velvet pivot]

-6 times -6?

oh

it's equal to negative 12

[late rose]

-6 * 6 = ?

[velvet pivot]

oh my god im so stupid XD

-36?

[late rose]

yes

18 = -36a

so a = ?

[velvet pivot]

so it's equal to -0.5?

[late rose]

yeah

[velvet pivot]

YAyyy tysm

[late rose]

you just had a silly arithmetic error, that's all

lemme read the second question

[velvet pivot]

Thanks! okay

[late rose]

read the second question very very carefully

and maybe try drawing it out

[velvet pivot]

At 2:30 PM, would the x-axis girl be at 2x+0.5 and the y axis girl be at x+1?

[lethal estuary]

Almost

Is the x-axis girl Kara or Lindsey?

[velvet pivot]

X is kara

[late rose]

and the question says kara moves what amount at 2.30?

[velvet pivot]

x amount of miles +1 mile right?

Wait

[late rose]

kara moves 2x at 2.00

[velvet pivot]

Yes

[lethal estuary]

Right now your sketch states that Kara moved an extra half a mile at 2:30, and Lindsey moved an extra mile

[velvet pivot]

when i was solving this for the first time, i just subtracted 1 mile from Lindsey and um i dont know

[velvet pivot]

Yes because in the question at 2:30 PM the girl that's length moved away from the camp North is represented by x + 1 mile

and the girl that moved west, Kara, is twice the distance away and also travelled 0.5 miles

oh i switched them around

[lethal estuary]

Yes

[velvet pivot]

XD

[lethal estuary]

That's what I was trying to get you to notice

[velvet pivot]

Ah

I assume if it were this way that my answer would be correct though?

[late rose]

yeah so basically you just need to work on reducing your careless mistakes

[velvet pivot]

Yayyy thanks a lot! I feel like I have a way better grasp on this now because it's Algebra 1b that im taking

[late rose]

are you allowed to use a calculator?

[velvet pivot]

Yes, but it takes me so long to solve questions so i just try to be fast which causes me to make mistakes

[late rose]

cause in the first question if you had just typed in (-2 - 4)(-2 + 8) into your calculator you would've just avoided that careless mistake entirely

[late rose]

ah okay

dont be afraid to go slow

[velvet pivot]

i was writing the stuff down, but when i got to this part i did the rest in my head so i thought -6+-6 = -12 XD

[velvet pivot]

Thanks!

.close

[late rose]

!done

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[velvet pivot]

.close

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[warm prairie]

hey helpers

[warm prairie]

[\int \frac{\ln(x)}{x - k},dx]

can this integral be solved?

Im trying

[soft zealot]

Bro

[warm prairie]

k is a constant

are you sure man

hold on

bro it cant be

[\int \frac{\ln(x)}{x - k},dx=\ln(x),\ln|x - k| - \int \frac{\ln|x - k|}{x},dx
]

[soft zealot]

Bro

[vital crag]

,w int log(x) / (x-k)

[warm prairie]

how tf do I even solve this

[soft zealot]

Results provided by WolframAlpha

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

Requested by rie.mann

[warm prairie]

smart

[vital crag]

https://mathworld.wolfram.com/Polylogarithm.html

[warm prairie]

Interesting

Def out of my league

fook it

thanks however

.close

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[keen wolf]

What is the equation of the line tangent to the curve at the given point? xsin2y=ycos2x at (pi/4, pi/2)

[keen wolf]

I know that you have to use implicit differentiation and apply the product rule and chain rule to solve it, but im not sure if im getting the correct answer and chatgpt is too stupid for this type of thing

[vital crag]

!show

[final saddle]

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

[keen wolf]

left side: 1(sin2y) + (cos2y dy/dx)(x)(2 dy/dx)
right side: dy/dx (cos2x) + (y)(-sin2y)(2)

right now, i am confused on whether or not i should have the dy/dx after cos2y after taking the derivative of sin2y in the product rule part of the right side, and im stuck on how to get the derivatives to all be on one side

[final saddle]

@keen wolf Has your question been resolved?

[loud sundial]

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

[soft zealot]

Civil Service Pigeon

[loud sundial]

isolating dy/dx is standard algebra

recall how you solve equations of the form ax+b=cx+d by getting all the terms with x on one side and the constant terms on the other

the idea is the exact same here

[keen wolf]

now im confused as to why cos2xdy/dx is crossed out (deemed to equal 0) and not included in the 2nd step of the solution

[deep condor]

sub x = pi/4 and see what happens :)

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[shrewd juniper]

I'm fairly certain my answer is right. Why isn't it taking it?

[shrewd juniper]

This is WeBWork

I tried the U instead of a comma and that didnt work but that still doesnt answer why the first one isnt working anyway

<@&286206848099549185>

[muted prairie]

!15m

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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.

[muted prairie]

I guess you could try closed brackets?

Would be a bit weird but thats the interval a set theorist would say it's increasing on

[shrewd juniper]

I'll try it

okay it worked

why would a set theorist say that

Well thank you so much

.close

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[tranquil pine]

How do I reflect

[vestal crest]

first, imagine or screencap/draw a line at wherever they told you to.
then, apply the reflection that way.

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

Huh

[vestal crest]

which part do you not understand? or perhaps you don't get what reflection is?

[tranquil pine]

Honestly none of it

[tranquil pine]

I never really understood it

And he hasn’t taught us about it in class but we got a test tomorrow

[vestal crest]

then this is the idea of reflection. suppose the green shape is the original and the red one is the image.

to get from green to red, draw a straight line, horizontal or vertical*, from one of the corners of the shape to the line of reflection (the "mirror").

then, extend the line by the same length (if you drew 2cm from the original point to the mirror, draw 2 more cm in the same direction).

do this for all of the points of the shape to get your reflected image.

[vestal crest]

*I assume you won't see slanted mirrors, but if you do, the line you draw should be perpendicular to the mirror (aka 90 degrees).

[vestal crest]

this step and the one above looks like this.

[tranquil pine]

So it would go over 2?

since its X=-2

[vestal crest]

for your case, the line x = -2 is here, and so that's your mirror.

[tranquil pine]

And since it’s negative it would go down?

[vestal crest]

no. it will go left. there is no downward movement necessary here.

[vestal crest]

your choice is D?

[tranquil pine]

So it wouldn’t be A?

[tranquil pine]

I guessed

[vestal crest]

A is definitely out.

[tranquil pine]

Wait would it be c since it’s reflective over the line you drew

[vestal crest]

D is close, but not the right image. the image is too close to the mirror.

[vestal crest]

C is correct! and that's exactly the right idea.

you want the image to be a reflection over the mirror.

[tranquil pine]

How would I do it ifs it’s Y instead of X

Like this

[vestal crest]

have you learnt about the equations of horizontal and vertical lines, actually?

[tranquil pine]

No

[vestal crest]

they will be useful here, but in case you haven't, y = {some number} is a horizontal line.

[vestal crest]

consider looking up on them.

[tranquil pine]

I will

[vestal crest]

anyway for the case of y = number, this is your situation.

[tranquil pine]

So y=3?

[vestal crest]

well, yeah, that's what you've been given.

[tranquil pine]

Would you draw the line on 3 then reflect over it?

[vestal crest]

yes.

that's exactly what a reflection is, isn't it?

[tranquil pine]

I think

[vestal crest]

it's in the name, so.

anyway, gotten the answer?

[tranquil pine]

No

How would I reflect

[vestal crest]

how did you do it for the first one? apply the same strategy here.

[tranquil pine]

Uhhhhh

The first was easier

I know it’s A but idk how I translated it

Or reflected

[vestal crest]

it's more or less the same idea here.

except that you're now reflecting upwards rather than leftwards because the mirror is above your object.

if in doubt, reflect each point separately.

[tranquil pine]

So go up three

[vestal crest]

not necessarily. see, point K is only one unit under the mirror. it doesn't need to go up three.

if point K is one unit under the mirror, then point K' (the image) will be one unit above the mirror. makes sense?

[tranquil pine]

Can you dumb that down

[vestal crest]

stand in front of a mirror. notice that you are as far from the mirror as your image seems to be. do you agree?

[tranquil pine]

Yes

[vestal crest]

if your hand is 3cm away from the mirror, the image's hand is also 3cm away from the mirror, just from the other side.

[vestal crest]

that's the idea.

since K is 1cm from the mirror, K' will be 1cm from the mirror as well, but from the other side.
since K was under the mirror, K' will be above, hence other side.

[tranquil pine]

Why couldn’t it be B?

[vestal crest]

draw a line at y = 3. you'll see why.

[tranquil pine]

Okay thank you

[vestal crest]

glad to have helped.

[tranquil pine]

This makes no sense bro

[vestal crest]

what makes no sense?

[tranquil pine]

Everything

I can’t do it

[vestal crest]

what do you mean, though? you did it for the first one after I drew the line for you, and you correctly identified that you need to reflect your object over the mirror.

[tranquil pine]

I tried to do two on my own and got them wrong

And don’t understand it

It’s stupid

[vestal crest]

since I don't understand your confusion, let me list some possibilities and you tell me which one you're confused about.

1. I don't know where y = 3 is.
2. I got y = 3, and drew the line, but I don't know how to reflect the object over.
3. I reflected the object over, but I got the wrong answer.
4. None of the above.

[tranquil pine]

I don’t understand anything

None of it makes sense

[vestal crest]

why does reflection not make sense, in your opinion?

you clearly identified in the first question that you have to reflect the object over the mirror to get the image, and identified the correct image with some help.

[tranquil pine]

Cause you basically did it from me

[vestal crest]

so one would presume that reflection already makes sense from the way you phrased your answer here.

[vestal crest]

in that case, let's reproduce what I did, but this time I'll only tell you what to do. you do it, and show each step.

step 1: draw a line at y = 3.

(hint: if you're on Windows, take a screenshot with Win + Shift + S, then draw on the screenshot.)

[tranquil pine]

Yeah but what could y=x mean and I did the same thing for y=2 and got it wrong

It’s all stupid

[vestal crest]

now you never said anything about y = x and y = 2. your question says y = 3.

[tranquil pine]

I know I tried to other that said that

[vestal crest]

also, have you learnt how to graph linear/straight-line equations before tackling this chapter?

[tranquil pine]

No

He hasn’t taught us anything

[vestal crest]

in that case, show what you did! I can't help what I can't see.

[tranquil pine]

Not even has to use demos

[tranquil pine]

How does y=x make sense

[vestal crest]

why not?

okay, maybe it doesn't make sense to you because you are not taught linear functions, in which case I urge you to study that first.

but in this case, y = x is the set of points where the x/y-coordinates are the same (like (0,0), (1,1), etc.)

[tranquil pine]

I have to know this by tomorrow

[vestal crest]

tomorrow or not, the basics are lacking.

anyway...

[vestal crest]

so you would expect to see a straight line slanted upwards (from bottom left to top right). just for this one, I'll draw it for you.

the image for this one will be located in the bottom right corner, so you'll have to move down and right this time around.

[tranquil pine]

So b would be on bottom?

[vestal crest]

point B would be the most bottom-right.

and yes, B will now be on the bottom of the shape.

[tranquil pine]

Then it would be in the 4th?

[vestal crest]

if you mean 4th quadrant, yes.

but the entire image shape would be in the 4th quadrant, save for point D' which lies on the x-axis.

[tranquil pine]

So it would be upside down in the 4th quadrant

[vestal crest]

it would be facing the right side in the 4th quadrant. (as in, BC will be on the right, and AD on the left.)

[tranquil pine]

So like flipped upside then turned

[vestal crest]

kind of, yes.

it'll look something like this.

[tranquil pine]

How do you do this so easily

I feel like a idiot

[vestal crest]

by first understanding what reflection does to an object, then understanding how to graph a line.
from there, it's drawing lines to connect the object to the mirror, and then extending them to the image.

[tranquil pine]

How do I do it without a graph

[vestal crest]

what do you mean?

[tranquil pine]

It just gives me the cords

[vestal crest]

the blue shape is your original shape.

technically you don't need the coords, unless you want to check the blue shape against the coordinates.

[tranquil pine]

How would I do it without a graph

[vestal crest]

again, I don't understand what you mean.

[tranquil pine]

I only can do it when I out a line with just the numbers idk what to do

Im sorry being so stupid

[vestal crest]

why are you so fixated on the coordinates? the answer options show you the original shape!

[vestal crest]

also, the line is given here!

the only reason you might want to refer to the coordinates is to check whether an option deliberately drew the original shape in the wrong place.

(of course, if you're gungho, you can draw the entire diagram yourself, original shape, mirror and image included.)

[tranquil pine]

So how do I get the answer if it’s just 1

[vestal crest]

it's x = 1.

draw in the line in every option and check whether the reflection makes sense.

alternatively, you can draw the diagram from scratch, but given how confused you are right now, that probably won't help.

a reminder that x = 1 is a vertical line.

actually wait a minute. they are generous enough to draw in the line for you!

you just need to check whether the line is correct, the original shape is in the correct place (if you want to) and if the image makes sense.

for you, though, I would recommend checking the original shape last, because it is the most tedious to check. check the line and the image first.

I will tell you right now that A is wrong in like eight different ways. (this is an exaggeration, if you can't tell.)

try the rest yourself.

[tranquil pine]

Im just gonna fail it this is way to complicated

[vestal crest]

if so, I bid you a good day, and you may close the channel when you're done, or leave it open if you decide to come back to it.

[tranquil pine]

.close

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[unique gale]

this is my problem. so far ive shown that the two sets for the constraints are bounded and closed, but now idk how to prove the minimum part.

[final saddle]

@unique gale Has your question been resolved?

[supple marten]

Not an answer, but a way to get the distance value: Given that the tangent to x^2+2y^2=1 at the closest point to a line is parallel (or equivalently the shortest distance is along a normal line out) the tangent at P being x x(P) + 2 y y(P) = 1, with the slopes matching when x(P)=2y(P), solving the system with the other equation being x(P)^2+2y(P)^2=1, we get (2/sqrt6,1/sqrt6) (and also the point with the tangent on the other side), the tangent there is (x+y)/sqrt2=sqrt3/2 which has distance 5sqrt2-sqrt3/2 or approx 6.2050... from (x+y)/sqrt2=5sqrt2 just x+y=10 written in unit covector form.

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[bleak elbow]

someone please help me I am so so tired

[bleak elbow]

i thought the answer was supposed to be 0.943 but that s wrong

[final saddle]

@bleak elbow Has your question been resolved?

[final saddle]

@bleak elbow Has your question been resolved?

[muted prairie]

Did you calculate P(X=12) instead?

,w nCr(20,12).6^8.4^12

[soft zealot]

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Requested by a_mog

[muted prairie]

,w sum from k=0 to 11 of (nCr(20,k).6^k.4^(20-k))

ok wtf is this

[soft zealot]

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Requested by a_mog

[muted prairie]

Idk why this is <1/2

I give up

Wait no I'm doing it backwards

,w sum from k=0 to 11 of (nCr(20,k).4^k.6^(20-k))

[soft zealot]

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[muted prairie]

that looks close to .943

,w sum from k=0 to 12 of (nCr(20,k).4^k.6^(20-k))

[soft zealot]

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Requested by a_mog

[muted prairie]

that looks like .979

idk I guess your teacher screwed up

they counted X=12 as a false positive

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[timber plume]

Hi, how do I know which formula sum to use? the Arithmetic or geometric?

[thorny canyon]

all of these are geometric, because a common factor is being multiplied each consecutive term

[timber plume]

how did u figure it out? because if you calculated all the terms, that would take time

[thorny canyon]

all the summands are written in form $ab^n$

[soft zealot]

Donkey

[craggy plank]

Let's take a for example, can you write the first three terms of it?

[timber plume]

2, 4, 8

[craggy plank]

yep

Do you notice anything?

[timber plume]

yeah I know this

the r=2

but this was not my confusion

[craggy plank]

what is your confusion?

[timber plume]

what are summands?

[thorny canyon]

things that are getting added

[timber plume]

I see

so if they are written with a power

its a geometric?

[thorny canyon]

its geometric if $\frac{a_{n+1}}{a_n}$ is constant for all values of n

[soft zealot]

Donkey

[thorny canyon]

where $a_n$ is the $n^{th}$ term of the sequence

[soft zealot]

Donkey

[thorny canyon]

this assumption is incomplete cuz agp such as $\sum_{i=1}^n(2i-1)b^i$ may also exist

[soft zealot]

Donkey

[timber plume]

I think I am lost

[timber plume]

I understand this

[timber plume]

not this

[timber plume]

not this

[thorny canyon]

here you say r=2 right?

[timber plume]

yes

[thorny canyon]

what's the definition of r?

[timber plume]

common ratio

[thorny canyon]

yeah

and how do you calculate common ratio?

[timber plume]

divide

by the previous term

[thorny canyon]

that's right

[thorny canyon]

and that's exactly what's written here

whenever you divide a term by its previous term, you always get same ratio

that's what a gp is

[timber plume]

oh i see

makes sense

[timber plume]

what about this?

[thorny canyon]

different concept altogether

its not an Ap or gp

[timber plume]

in what cases?

[thorny canyon]

yk in ap, difference bw consecutive terms is constant

and in gp ratio bw consecutive terms is constant

[timber plume]

yes

thats what common means

[thorny canyon]

in agp see the (2i-1) part behaves like an ap, the $b^i$ part behaves like a gp

[soft zealot]

Donkey

[thorny canyon]

so its a mix of both, so to speak

sequence looks something like b, $3b^2$, $5b^3$, $7b^4$ and so on

[soft zealot]

Donkey

[timber plume]

whats agp? arithmetic geometric progression?

[thorny canyon]

yeah

[timber plume]

isnt it just

AP

and GP

[thorny canyon]

its a sequence where one part resembles ap, another resembles gp

[timber plume]

I see

thanks

.close

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[vague anchor]

24th please

[severe verge]

,rotate

[vague anchor]

<@&286206848099549185>

[severe verge]

Do you know how to find the coordinates of the incenter of a triangle?

[vague anchor]

Equidistant from sides?

[severe verge]

yes, the intersection of 3 angle bisectors

there's a quick formula for it

[vague anchor]

🤔

ax1....../a+b+c

For x coordinate

[severe verge]

seems like you're getting the idea...

for x coordinate it is (axA + bxB + cxC) / (a+b+c)

likewise for y

[vague anchor]

Yea

[severe verge]

you might be wondering why this is important

[vague anchor]

Yea

I don't see use of it

Here

[severe verge]

here we have ∆ABC

[vague anchor]

Yes

[severe verge]

D, E and F are perpendicular feets

[vague anchor]

H is orthocentre

[severe verge]

and we know its coordinates

yep

[vague anchor]

Go ahead

[severe verge]

H is actually the incenter of ∆DEF

[vague anchor]

Oh wow idk that

[severe verge]

you can easily prove that DH, EH and FH are angle bisectors in ∆DEF

this is true for any triangle

[vague anchor]

Yea agreed

Go ahead

[severe verge]

oh wait a second

I think this is not true for obtuse triangles

[vague anchor]

Yea

Nah it is ig

[severe verge]

huh

in that case one of the vertices we want is the incenter

[vague anchor]

??

[severe verge]

interesting

[severe verge]

anyways getting back to this acute triangle case

H is the incenter of ∆DEF so you can easily find its coordinates

then you just find the coordinates of A so that A lies on DH and AE is perpendicular to HE

likewise for other points

[vague anchor]

Th

.close

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[wicked crystal]

hey

[wicked crystal]

.close

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[proper dagger]

[tired walrus]

[latent dragon]

yeah…

[glossy zephyr]

fire stuff

[severe verge]

hmm scam service...

[craggy plank]

Indeed

[latent dragon]

5 greens 1 blue. 👀

[bleak granite]

nah

[craggy plank]

nah, I'm purple

[latent dragon]

forgot about that.

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[vague anchor]

32
,rccw

[proper dagger]

,rccw

[final saddle]

@vague anchor Has your question been resolved?

[vague anchor]

<@&286206848099549185>

[barren schooner]

?

[vague anchor]

I want help with the question

[barren schooner]

Any questions?

[vague anchor]

This 32

[lunar jewel]

Do you know that for an Equilateral triangle...
Orthocentre=Circumcentre=Centroid

[vague anchor]

Yea

[lunar jewel]

Yea so have used it?

Hint:- H=(A+B+C)/3

||Then you may write the co-ordinates of H explicitly||

Is the answer 8

@vague anchor

[vague anchor]

Bruh like i don't know that

Gimme some better stuff

[vague anchor]

Yea

[lunar jewel]

Kindly reinstate what you want to convey I did not get you

[vague anchor]

I Obv know that

[vague anchor]

This

Gimme further hints

[lunar jewel]

Centroid

The Coordinates of Centroid

The sum of co-ordinates of vertices of triangle divided by 3 (As the segments are divided in the ratio of 2:1)

[vague anchor]

Oh god the hell u mean

I know that

[lunar jewel]

Then why were you unable to solve the question

[vague anchor]

Gimme vertices of A, B and C

Who did u get that

There would be equations to solve these 3 variable problem how shall I get it

@lunar jewel ??

[lunar jewel]

I hinted you

[vague anchor]

Bruh

That was a hint

I knew that

[lunar jewel]

Try this

[vague anchor]

Oh god

[lunar jewel]

Wut

[vague anchor]

Just leave and stop fucking me here

.close

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[shadow gyro]

I did this statistics question and I got all three of my tries wrong

[shadow gyro]

I did E, H, and C for d)

does anyone know what I should've done to get it right?

[final saddle]

@shadow gyro Has your question been resolved?

[coarse loom]

Hi i have a question so when iam proving the limit of the sequence i can keep increasing the value of the fraction

[cursive bough]

you reject your null hypothesis when the p value is less than α = 0.01, not greater

[cursive bough]

this channel is occupied, you need to start a new help channel

[coarse loom]

what is the system of this server i dont get it

how to start

[shadow gyro]

so it would've been D?

[cursive bough]

#❓how-to-get-help

[cursive bough]

yes

wait

no, neither of them are correct

[shadow gyro]

hm

[cursive bough]

they both say the same thing but in different ways

also you never compare the critical value to the p-value, they are measuring different things

you either compare the test statistic to the critical value or you compare the p value to α

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[sturdy badge]

SOMEONE explain this plEASE

[thorny canyon]

you can use ratio test for convergence for such stuff

[sturdy badge]

wdym

i thought that was only for series

and not sequences

[thorny canyon]

formally yeah

[sturdy badge]

can you explain

how id do that

[thorny canyon]

well it's a standard result that if $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L<1$ then $a_n\to0$

[soft zealot]

Donkey

[thorny canyon]

or you could js say $0<a_n<2x\frac{2}{3}x\frac{2}{4}x\frac{2}{5}x\frac{2}{6}x\frac{2}{7}....x\frac{2}{n}\to0$ as $n\to\infty$

[soft zealot]

Donkey

[thorny canyon]

x as in multiplication 😭

[sturdy badge]

.close

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[balmy forge]

i dont recognize this from my notes so im not sure where to start

[thorny canyon]

yk polynomial long division?

[balmy forge]

i do

[thorny canyon]

that's the question basically

divide the polynomial by k-2

the quadratic you get is your answer

[balmy forge]

alright alright thanks

.close

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[late rose]

there's a faster way, by comparing coefficients like so: Let
$$k^3 - k^2 -k - 2 = (k - 2)(k^2 + bx + 1)$$
Whence, compare coefficients of $k^2$

[final saddle]

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[warm flame]

Hi, is this correct?

[coarse aurora]

You should label your axes

[coarse aurora]

The line y = 12 is not correct, the parabola is also not correct. If you labelled the axes the other way then it would be fine (besides y = a which will not be perpendicular to y = 12)

[warm flame]

True, i'm going to continue studying this, ty!

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[timber plume]

is my answer correct for q8?

[gloomy river]

Yeah, that's correct!

[atomic moon]

Yep

[timber plume]

thank you!

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[wheat oriole]

The answer key says this should be θ(n) but I don't get it why shouldn't it be θ(nlgn) θ(x+y) is basically θ(n) and the recursion runs lg n times

[wheat oriole]

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[final saddle]

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[gritty flume]

Can I get help w number b1)

[thorny canyon]

,rccw

[thorny canyon]

uh

multiple lhs by $\frac{\sqrt{2}}{\sqrt{2}}$ ig

[soft zealot]

Donkey

[gritty flume]

I don’t understand

[thorny canyon]

its given that you have to use $\frac{\sqrt{a}}{\sqrt{a}}=1$

[soft zealot]

Donkey

[thorny canyon]

and we know that multiplying anything by 1 doesn't change its value

so lhs x 1 = lhs

[gritty flume]

i still don’t get it

why is it root2 over 2

this must be dome without a calculator btw

[thorny canyon]

what is 1/root(2) times root(2)/root(2)

[gritty flume]

oh ic

[thorny canyon]

well yeah

[gritty flume]

then how do i solve b2)

[thorny canyon]

I take it you remember values of cos(45), tan(30), sin(60) and tan(60)?

[gritty flume]

yes

[thorny canyon]

so b2) (i) is just asking you the same question as b1)

b2) (ii) is asking you to prove 1/root(3) = root(3)/3

and so on

[gritty flume]

i still don’t get it

what do i multiply the 1/2 by

for b2) question i

[thorny canyon]

whats the value of cos45?

[gritty flume]

1/2

[thorny canyon]

no

[gritty flume]

huh

[thorny canyon]

1/2 is cos60

[gritty flume]

1/root 2

[thorny canyon]

right

so you gotta prove 1/root(2) = root(2)/2

[gritty flume]

yuh

[thorny canyon]

multiply 1/root(2) times root(2)/root(2)

what do you get?

[gritty flume]

is tht for all of them

i kind of get it

but what does it have to be root2/root2

[thorny canyon]

because in the question it wants you to use root(a)/root(a) = 1

try b2) (ii)

what do you multiply to tan60 to prove it?

[gritty flume]

root3/root3?

[thorny canyon]

correct

[gritty flume]

ic

ty

how do i do an inverse without a calculator

[rugged merlin]

Inverse of a function?

[gritty flume]

yuh

The c)

Please

[rugged merlin]

,rccw

[rugged merlin]

Do you know your trig table

[gritty flume]

I don’t think so

[rugged merlin]

Well you should

[gritty flume]

What’s it about

[rugged merlin]

This

U gotta remember those

[gritty flume]

Dang🥀

[rugged merlin]

Or if you dont want to remember derive them via the 45-45-90 and 30-60-90 triangles

[rugged merlin]

Anyways for the first one for example

sin(x) = 1/2

[rugged merlin]

You look at this table and see that it is x = 30 degrees

[gritty flume]

Dang

I gotta remember all the too

[rugged merlin]

Ye

[gritty flume]

What if I just use a calculator

[rugged merlin]

Probably not a good idea

Bcuz they probably wont allow u to use one in the exam

And you wont remember by then

[gritty flume]

💔

[tired walrus]

this is like the times tables for trig

[gritty flume]

why is there a timetable for this

[tired walrus]

i didn't mean timetable like schedule

i meant times tables like multiplication table

[gritty flume]

oh

[gritty flume]

what is csc, sec and cot

[rugged merlin]

The multiplicative inverses of sin, cos, and tan, respectively

[tired walrus]

csc(x) = 1/sin(x)

sec(x) = 1/cos(x)

and cot(x) = 1/tan(x)

[rugged merlin]

Ann beat me to the punch 😔 (damn typing on phone)

[gritty flume]

Wait what’s the inverse?

[gritty flume]

I don’t get Thai

[rugged merlin]

What ann said

Like the fractions

[gritty flume]

so is it for only finding the inverse or are those the general terms

[rugged merlin]

Idk what you mean by thay

[gritty flume]

can u explain everything again

[rugged merlin]

You know sin cos and tan yes?

[gritty flume]

yes

[rugged merlin]

Ok so

Csc is quite literally just

csc(x) = 1/sin(x)

You just flip it

So if sin(x) = 1

Then csc(x) = 1

If sin(x) = 1/2

Then csc(x) = 2

[gritty flume]

so csc= hypp/opp?

[rugged merlin]

Just flip the fraction if u wanna convert to rhe other

[gritty flume]

like tht

[rugged merlin]

Yuh good job

Same applies for cos-sec and tan-cot

[gritty flume]

tysm

[rugged merlin]

So

cos(x) = 1/2

Can u figure that our from the table

[gritty flume]

no

did you already flip the 1/2

[rugged merlin]

I flipped nothing

[rugged merlin]

Where is 1/2 here on the row of cos

[gritty flume]

60

[rugged merlin]

Ok

There u go

Do that for the rest

[gritty flume]

I’m still confused

that’s for the normal ones right

[compact saddle]

you could just remember the table for sine and cosine(which is the sine table but goes backwards) and get the other values using tan=sin/cos sec=1/cos, csc=1/sin in the beginning

though you will remember better as you practice the questions
atleast thats what i did

also helps to know that the value of sine keeps increasing from 0 to 90 while that of cos keeps decreasing

[gritty flume]

Ty

[compact saddle]

also remembering the value of sin and cos at 45 degrees as √2/2 might it more obvious that the value is increasing for sine as decreasing for cos and also make it easier to remember in the beginning since everything would be divided by 2 then

(you can verify 1/√2=√2/2 by multiplying √2 on numerator and denominator)

though memorise whichever one feels better to you

[gritty flume]

How do I do this please

Is it just the normal formula

[odd seal]

,rccw

[odd seal]

yes, directly use the definitions of the quantities

[gritty flume]

Ty

I got the b2) wrong

Could someone help me

[final saddle]

@gritty flume Has your question been resolved?

[compact saddle]

i dont see b2) in the image u sent earlier

[compact saddle]

,rccw

[compact saddle]

i dont see how you got sin 30=5.8/h

[gritty flume]

I run out of books to use

[compact saddle]

uhh these look like answers to questions other than the Q 2) you sent above

[gritty flume]

Wait which is it again

I don’t rlly remember

[compact saddle]

oh nvm

it is the saame question mb

were you asked to use decimal approximations

cus usually you just use root 3 as is

[gritty flume]

To one decimal point

[compact saddle]

i see

was the answer supposed to be

around 44.4 degrees?

[gritty flume]

Yes

But u got 44.1

[compact saddle]

the problem is that you used a approximation of root 3 to calculate the length of BD

also it can clearly be seen that BD eqials 10 since BDEF is a rectangle but your calculations resulted in 10.0smth cuz root 3 was approximated

[gritty flume]

I bc it another questions pls

Ig it

I got

[compact saddle]

o

sure

whats the next question?

[gritty flume]

Question b

Do you think the work must be done in the first or second wuadriant

Bc y and x are positive

And sin is greater than 0

[compact saddle]

they gave the value of tan as 5/12

can you deduce which quadrant it is from both tan=5/12 and sin>0?

[gritty flume]

It’s either the first or second

But I think the first

Bc everything is positive

[compact saddle]

thats ony using the fact that sin>0

[gritty flume]

The axis are positive

[compact saddle]

yes, since tan is negative in the second quadrant, you have to go in the first

[gritty flume]

tysm

But would there be a specific way to draw the triangle

[compact saddle]

[gritty flume]

.

[compact saddle]

what to you mean by specific way?

[gritty flume]

How will the triangle be drawn on the quadrants

[compact saddle]

yeah there is a specific way if your using the coordinate system

to represent it

[gritty flume]

Oh ok

[compact saddle]

the hypetoneuse has to make the angle theta being considered with the positive x axis

[gritty flume]

Scum

Wdym

Could you like draw it

[compact saddle]

https://www.youtube.com/shorts/3yT1ndjaoVE

this might make it easier to understand why it is that specific way

[gritty flume]

Ty

[compact saddle]

i think you already saw smth like this before since you know about the values being negatve in different coordinates

[compact saddle]

np!

[compact saddle]

btw, its not that you will get a different answer if you drew it a different way, its just that (rcosθ,rsinθ) being the coordinates of a line with length r starting from the origin making a angle theta with the positive x axis is a neat fact

and is mostly what "representing" the trig functions in a coordinate system would conventionally mean