help-23

[lost cove]

In triangle ABC, angle A = 65 degrees and AC = BC = 12cm. Determine the angles of the triangle

[thin bridge]

!status

[safe radish]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[lost cove]

1. I don't know where to begin

[thin bridge]

have you drawn a diagram

[lost cove]

What is a diagram

[thin bridge]

picture

sketch

[lost cove]

Yeah

[thin bridge]

visual representation of the problem

so you have done something

if you've done something, 1 isn't an appropriate response

show what you've done

[thin bridge]

knowing that two of the sides have the same length of 12,
can you identify what type of triangle you have

[lost cove]

I don't know man

[thin bridge]

have you heard of terms like
scalene, isosceles, equilateral?

[lost cove]

doesn't sound familiar

[thin bridge]

i would recommend looking them up

[lost cove]

Yeah i have seen that type of stuff before

[torpid shale]

what type of triangle do you think you have

[lost cove]

I'm guessing iscoceles

[torpid shale]

yep

so that means two equal angles as well

angle abc = angle acb

[lost cove]

you're right

65+2x=180

[torpid shale]

yes

very nice

[lost cove]

For which angles v in a right triangle does sin v < cos v apply?

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[strong flower]

Pls help

[strong flower]

Question 6.b… 6.a answer is 7.81

[thin bridge]

what have you tried?

[strong flower]

Oh wait sorry the answe for 6a is 7.81

I’m having problem with 6b and c

[thin bridge]

have you made any attempt to solve b and c

[strong flower]

I’ve tried to find the degree but it’s doesn’t make sense since it needs at least two numbers to find the theta

Yes

The answers are wrong

[thin bridge]

show work

[thin bridge]

what you wrote down for b) was just converting 90° to radians which isn't what they're asking for nor is it needed

[strong flower]

Yep

I know but idk what else to do

[thin bridge]

there are multiple approaches

but from part a)

its already implied that you know the perpendicular bisector theorem
and that you formed a right triangle to be able to apply pythag

then use inverse trig to determine some angles which will ultimately lead you to the angle they want

[strong flower]

I haven’t learn this

Or inverse trigonometry

[thin bridge]

not sure how they're expecting you to do this without inverse trig

[strong flower]

This is the first topic before trigonometry

I donno either

Second subtopic in new chapter

Circular measure

[thin bridge]

if you don't know trig, skip for now

[strong flower]

I’ll ask my teacher then

Sorry to bother you

Thanks

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[lone lagoon]

hi i solved but can someone sketch the derivative diagram? Did it without understanding

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@lone lagoon Has your question been resolved?

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[wary mountain]

can i get some help

[wary mountain]

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[safe radish]

Please don't occupy multiple help channels.

[charred grove]

Is the answer here A? Seems too simple

[compact ferry]

yeah its A

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@charred grove Has your question been resolved?

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[steel remnant]

In set theory:
can I say that if A \union {1,2} = B \union {1,2}
Then A = B?

[pallid nymph]

no

Let B = A \union {1,2}

and A not have 1 or 2

[steel remnant]

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[true estuary]

why would the cooridnate matrix for part b not be this?

[granite idol]

multiply it by each of the standard basis vectors, I think the answer should be the transpose but I'm not 100% sure

I get tripped up by change of basis problems too

[true estuary]

yh mb, the matrix should acc be the columns of the ordered basis

these here

[granite idol]

it might be the inverse... TBH I don't remember the direction to go in...

[true estuary]

forgot i had an MS for this, just checked and its this

so yh its the columns of the ordered basis

[granite idol]

ok, so it was the transpose of what you originally had

[true estuary]

ye

cool

cheers mate

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[tall hare]

i tried simplifying the expression before reciprocating the resulting fraction but that was also wrong

[wet agate]

I thinnk the way you reciprocaled it is wrong

[tall hare]

how do i do it

[wet agate]

You can first find 1 value in 1 fraction for sin(a -b) then reciprocal it

[tall hare]

thats what i did but it said it was wrong

originally what i did

this it my first attempt

im guessing my rationalization of the denominator is wrong

^

[wet agate]

You have to mulitply the numberator and denomiator by 8 root 35 + 15 i think

[tall hare]

ok ill try that

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[steep gust]

why tan^-1(tanx) is x

[lean otter]

Check .46

I can send you this pdf if you want

[steep gust]

is all? trigometric funcition?

[lean otter]

Yeah, just a very particular formula

[steep gust]

okay send pdf

is this also?

[lean otter]

Oh, that goes beyond me, sorry.

I think it's just a pool of general formulas.

And I'm no trig expert.

[steep gust]

Where did you get those trigonometric functions?

[lean otter]

The school teacher gave it to us

It's from a romanian university or something.

[steep gust]

you cant pass the pdf?

[lean otter]

I sent it already in your DMs

[steep gust]

i see thx

[lean otter]

👍

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[shy hornet]

I think it is list 4

[floral osprey]

it says pulling 2 slips out

[shy hornet]

Yeah

Oh wait is it list 2 @floral osprey

[floral osprey]

seemslikeagoodguesstome

[shy hornet]

Ok

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[empty gyro]

I would remove the xy term using a rotation and convert to a standard conic form

I would also try parameters like (t,0), (0,t), and (t,t). If they're not all equal, then the limit does not exist

Yeah but here (0, t) and (t, t) do not

Google rotated conics

What second example?

Oh. No that won't work

Just creates more headache

Why are you trying to find this limit

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[timid goblet]

Check my answers for 8 and 9 and help me with 7

<@&286206848099549185>

<@&286206848099549185>….?

[green bane]

,rotate

[timid goblet]

Hi

@green bane

[green bane]

8 seems good

[timid goblet]

Mhm

[green bane]

9 check it again

[timid goblet]

I dont get it!! Whatttttt how is wrong

[green bane]

A is negative

$f(x)=f(−x)$ for even function

[flat frigate]

bacchess

[timid goblet]

Oh

Ok

..

Where u go dawg

@green bane

[green bane]

wha

[timid goblet]

I Need help wit 9

[timid goblet]

-50 still fits I to quadrant 4

[green bane]

A

[timid goblet]

Why my dumbass thought that A was positive

Holy shit im blind

[green bane]

:/

happens

[timid goblet]

So 7 I still don’t get too sorry

Can you walk me through #7?

[green bane]

lets do an example

f(x) = x^2

you there?

[timid goblet]

Yup

I’m here!

[green bane]

hmm what does adding do to a function

[timid goblet]

Adding makes the function a positive function?

[green bane]

i mean to the graph

[timid goblet]

Oh

The function Is going upwards?

[green bane]

so if it is even

does moving upwards change it

[timid goblet]

No

[green bane]

if you add something in the function what does it do

like uh $(x+1)^2$

[flat frigate]

bacchess

[timid goblet]

Idk

[green bane]

well is (x+1)^2 a even function?

[timid goblet]

Bo

No*

[green bane]

so adding in a function makes it not even

[timid goblet]

Oh

[green bane]

now

solve

[safe radish]

@timid goblet Has your question been resolved?

[safe radish]

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[charred dagger]

I have no idea how to answer questions 3 and 4 on this paper like I don't even know where to start and I can't find any explanations online 😭. Very much appreciated if someone could point me in the right direction! I hate Exponential Functions‼️

[charred dagger]

that picture is blurry asf im so sorry

[slate thorn]

I don't remember exactly but I'm pretty sure you just set y equal to 800k

[charred dagger]

is it rly that simple?

[slate thorn]

Yeah think abt it

I'm kind of blanking so it's hard to think of an example

[charred dagger]

wait im sorry where is y

[slate thorn]

Do yk the formula for exponential growth/decay

[charred dagger]

OMG

yes hold on

[slate thorn]

I think I can figure it out rn but I seriously just forget

[charred dagger]

it's okay

hold on im looking at my notes rq

[slate thorn]

y=a(1+r)^t

?

[charred dagger]

no I've never seen that before 😭

so is that what I'm supposed to be using?

[slate thorn]

Can yiu look through your notes and find an equation that models growth or decay for something

[charred dagger]

the one I have is this: f(x) = a*b^x

[charred dagger]

yes one sec

[slate thorn]

Oh b is just 1+r

[charred dagger]

these notes are from before spring break and I forgot like everything sorry

[slate thorn]

It's okay

[charred dagger]

thx!

but what's r supposed to be?

[slate thorn]

But a is the base right

R is the rate of change

[charred dagger]

i'm not sure

what I have written down says: a is a nonzero, b is a positive

i have an example of the problem that I did but I have no idea what's happening in it do you want to see it

[slate thorn]

Like if it goes up 30% each time you it would be 1+0.30

[charred dagger]

is that growth

[slate thorn]

Yes

So if you think about it

Sorry I'm on my phone so I'm a bit limited

Like let's say we start at the number 2

That will be a

[charred dagger]

it's okay i rly appreciate you trying to help

[slate thorn]

Here the starting number would be?

[charred dagger]

for texas or ohio?

ill do texas so 465,648

[slate thorn]

What does the question ask

[charred dagger]

resending so you dont have to keep scrolling

oh when it surpasses 800k

[slate thorn]

Tysm

You can just find when it does hit 800k

[charred dagger]

using the formula from earlier?

[slate thorn]

Oh I see this is a bit complicated

[charred dagger]

that makes me so happy i thought i was stupid LOL

[slate thorn]

Do you know if you did 1 and 2 right

[charred dagger]

No I don't know I was just guessing that's what it meant

it could be totally wrong

[slate thorn]

Yeah thats where the issues are I think

[charred dagger]

okay so im going to get rid of that

[slate thorn]

It says what is the growth factor right

[charred dagger]

yes determine the base and growth factor

[slate thorn]

So that would mean we're multiplying the base by a certain number to get to 931830

[charred dagger]

what is the base

or like how do i determine that

[slate thorn]

Base is just the starting number

[charred dagger]

so it is 465648?

just making sure

lmk if you want me to resend the doc

[slate thorn]

Yes

[charred dagger]

ok thx

[charred dagger]

and is the certain number the growth?

[slate thorn]

Well to find the rate is actually quite simple if we just use the formula j showed before

[charred dagger]

y=a(1+r)^t

[slate thorn]

Yess

[charred dagger]

ive never seen it before should i look up how to set it up

or is it easy to explain

[slate thorn]

So t represents the amount of time that has passed

[charred dagger]

!

so 25!

[slate thorn]

For now we can just memorize but it would be useful to understand where it derives from

Yes

And now we only have 1 unknown right

[charred dagger]

👍

is that r?

the unknown

[slate thorn]

Y would be the population given from the formula right

So if we want to find r at 25 years

931830=a(1+r)^t

Can you fill out a and t

[charred dagger]

t is 25 right?

[slate thorn]

Yes

Then a is

[charred dagger]

the base?

im not entirely sure

[slate thorn]

Starting number

[charred dagger]

465648?

[slate thorn]

Yes

[charred dagger]

WOOOO

[slate thorn]

Okay have you learned about logarithms ýet

[charred dagger]

yes I have

i have a better grasp on that

[slate thorn]

Okay then next part should be easy right

To find r

[charred dagger]

ill try it but can you tell me if I do it wrong

this is a logarithm?

[slate thorn]

931830=465648(1+r)^25

Yes but show your work I wouldn't be able to get an exact value without a calculator rn

Also if you want you can set b = to 1+b

[charred dagger]

how does this work?

[slate thorn]

A little bit

[charred dagger]

lol its okay i found an online one

[slate thorn]

1+r**

[charred dagger]

ahh okay

one second i have to be afk ill be back in a second

[slate thorn]

Np

[charred dagger]

okay i have a question on how to solve

do i devide both sides by 465748

*divide 😭

[slate thorn]

Yes it would make it simpler

[charred dagger]

okay just making sure

okay so

when i divide them should i be using 2.001146789 or just 1745/872

i feel like the latter would be easier but idk

[slate thorn]

Latter

[charred dagger]

😁 👍

and now i do the square root of both sides right

hold on ill send a pic

[slate thorn]

You would take the log of both sides

[charred dagger]

i dont understand

[slate thorn]

Yes that works too

[charred dagger]

oh okay great

so did we find r?

[slate thorn]

Yes

[charred dagger]

that's the growth right?

[slate thorn]

That would be the factor for growth

Do the same for the next part

[charred dagger]

YIPPEEE

okay will do give me one sec

sorry for the bad picture but heres what i got

[slate thorn]

That's good

[charred dagger]

HOORAY

tysm i rly appreciate this

you dont have to keep helping me you've done more than enough but is there a formula or something for 3 and 4?

also whatever you just taught me we did NOT learn it that way so thank you for helping me understand

[slate thorn]

Now we're trying to find t

800000=a(1+r)^t

[charred dagger]

hmm okay

and we already have r and a right

OH this should be easy then!

wait nvm not easy im about to be stuck in a second

why is the +1 even there they just cancel out 🤨

am i on the right track

pretend the t is there

im going to try to do 3 and 4 and wait for u to come back

[slate thorn]

@charred dagger

[charred dagger]

hey

[slate thorn]

Wdym they just cancel out

[charred dagger]

there was a +1 and then the -1 so they just cancel out right?

[slate thorn]

Oh but r isn't 1+r right

[charred dagger]

not r was a big number -1

so when you plug in that number for the r the 1's cancelled out

[slate thorn]

Oh yeah they would

You need to use logarithms for this btw

[charred dagger]

one sec sry

[safe radish]

@charred dagger Has your question been resolved?

[charred dagger]

@slate thorn TYSM FOR HELPING ME but i have to get off of my computer ill figure the rest out from here!!!

[safe radish]

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[dusk shale]

I don't know what I did wrong on this question. Please help/

[frank quartz]

so, we have a circle with a radius 85, we know the point a
to find point b, we need to make a right triangle

[dusk shale]

right.

[frank quartz]

that angle is 7pi/2

do you remember how to find the x and y on a unit circle from an angle, with the right triangle

[dusk shale]

isn't it x^2 + y^2 = 1?

[frank quartz]

yeah, but we want a formula of x from an angle

and y

[dusk shale]

I'm not really sure...

no.

[frank quartz]

one sec

[dusk shale]

👍

[frank quartz]

would you agree that the length of the right triangle is the x coordinate and the height is the y coordinate?

[dusk shale]

yes.

[frank quartz]

we have a right triangle, so we can use soh cah toa

[dusk shale]

right.

[frank quartz]

but, we need to remember sin(a)=o/h, then h*sin(a)=o

what is h?

[dusk shale]

height?

hypotenuse sorry

I was thinking about something else.

[frank quartz]

right, what is the length of the hypotenuse

[dusk shale]

in the equation you posted? h?

[frank quartz]

its the radius of the circle, right?

[dusk shale]

yes

[frank quartz]

okay, so we know that x=r*sin(a)

[dusk shale]

if r = 85 in the original problem, H = 85 right?

[frank quartz]

with a the angle to the positive x axis

[frank quartz]

exactly

[dusk shale]

okay.

so would 100 be the height?

?\

[frank quartz]

so, looking at B, our height y=r*sin(7pi/12) to the center, but we need to add how high the center is to get the height from the ground

[frank quartz]

i made a typo here, i meant y=r*sin(a)

[dusk shale]

we would add 100 right (as it's height)?

[frank quartz]

right

[dusk shale]

so for the equation of B what would that be?

100 + 85sin(7pi/12) = ?feet

[frank quartz]

you almost got it right, you added 85 instead on the image you sent

[dusk shale]

what did I do wrong in my initial problem?

[frank quartz]

you did
B=85 + 85sin(7pi/12)
the correct answer was
B=100 + 85sin(7pi/12)

[dusk shale]

okay.

how about A?

[frank quartz]

same thing for all of them

you need to add 100 instead of 85

[dusk shale]

but the base equations are right?

[frank quartz]

yeah, the idea is right

these equations assume the bottom of the ferris wheel is on the ground, but we know its a little off the ground

[dusk shale]

right

so

A = 100 + 85sin(0) = 85 B = 100 + 85sin(7pi/12) = 182 C = 100 + 85sin(5pi/4) = 50?

?

[frank quartz]

sorry, jumping around alot feel free to @ me, yeah

[dusk shale]

okay. thanks.

@frank quartz

[frank quartz]

okay

so clock times can be thought of as angles

60 possible minutes, whereas we have 2pi radians, we can think of it this way because 60mins=0mins

[dusk shale]

okay.

[frank quartz]

then, to find the angle we do 2pi radians = 60 minutes

we want radians, so we divide 60, 1 minute = 2pi/60 radians

now, we can multiply 1 by our minute, say 30 minutes = 30 minutes * 1 minute = 2pi*30/60 radians

which would simplify to 30 minutes = pi radians

just remember that in clock times, 0 minutes is directly upwards instead of rightwards

so to get to the triangle we talked about last time we need the angle to be = previous angle + pi/2

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[candid trench]

I can't figure out how to prove these. This is the start of my attempt at proving c. We just learned about induction and I don't completely understand it but I think I'm supposed to be using it?

[candid trench]

This was my attempt at b without induction

[brave edge]

Okay first of all

A(1,n) is a number

P(n) is a proposition

P(n)=A(1,n) doesn't make sense

By writing it correctly, you will probably help yourself at finding how to proceed for the c

[candid trench]

Will it make sense if I use ≡?

[brave edge]

More like ":"

Can you vc ?

[brave edge]

"Proof by example" is not proof

[candid trench]

I can't right now

That makes sense

[brave edge]

And can you just hear me while being mute ?

[candid trench]

I can in a few minutes

[brave edge]

Am waiting

[candid trench]

I'm sorry, it's taking longer than I thought it would

[brave edge]

I guess that will be the last thing before sleeping

Okay I won't be able to stay

Good luck !

[candid trench]

I was about to say I could now. Sorry.
Thank you!

<@&286206848099549185>

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@candid trench Has your question been resolved?

[safe radish]

@candid trench Has your question been resolved?

[queen ermine]

Hahaha

[gusty inlet]

@candid trench

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[oblique meteor]

anyone know prolog well?
this is my question, would you have any idea how to implement any of the features?

[pastel verge]

That needs to be clear

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[lean otter]

got no idea what to do ive revised my notes and i still have no clue

whats your definition of a "one-to-one function"

what does that mean

uh

non repeating inputs and outputs

uh okay sure

[random star]

bijective?

[lean otter]

what is the inverse?

no idea💀

when you switch each value?

<@&286206848099549185>

[random star]

think of functions as mapping between two sets

you have set A and set B

[lean otter]

mhm

[random star]

a function is literally an expression that maps en element from A (domain) to B (codomain)

"an arrow" literally

the inverse function is an arrow that maps an element from B to A

[lean otter]

so the answer of the opposite of the given value?

g(5)=-8 would be g^-1(5)=8?

[patent tangle]

no

if g(5) = -8 then g^-1(-8) = 5

[lean otter]

ah i see

[patent tangle]

the function and inverse function kinda cancel

[lean otter]

you swap the values

[patent tangle]

like g^-1(g(x)) = x

[random star]

this is your case

[patent tangle]

yeah i think mapping it is easier to understand

[lean otter]

and g(8)=2 would be g^-1(2)=8

[random star]

exactly

[patent tangle]

yes

[lean otter]

thats all?

[random star]

i wonder why they don't teach functions like this in HS. We had a logic/discreet math course in uni and it's been the most useful course I've ever had

[patent tangle]

based on what you have yes, we cannot say anything about the first 2 choices because we don't know if the function is odd or even imo

[lean otter]

i see

thank you

how do i close this

[patent tangle]

.close

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[frosty granite]

Is substitution the best way

[safe radish]

@frosty granite Has your question been resolved?

[wind agate]

u = t^5 -> du = 5t^4 dt would make things a bit easier maybe

[frosty granite]

k thanks

.close

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[frosty granite]

is that another way of checking for point of inflection?

[frosty granite]

finding third derivative

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[little smelt]

You can find without 3rd derivative

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[kind otter]

Hello, Im trying to prove this modul. I know how to do questions that dont have an extra element like "-13" but I don't understand how to do this one.

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@kind otter Has your question been resolved?

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@kind otter Has your question been resolved?

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@kind otter Has your question been resolved?

[lean otter]

You can add 64 in both sides (and remember that 64 ≡ 0. In that case we have -13≡ 64-13≡ 51

[kind otter]

Alright thanks

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[peak crane]

how do i find the lcd of x-1 and 1-x

[stoic dune]

Note that 1 - x = -(x - 1)

[peak crane]

yea

[stoic dune]

So x - 1 divides both

[peak crane]

-1?

[stoic dune]

(1 - x) / (x - 1) = -1

[peak crane]

or is it 0

[peak crane]

yea

so its -1?

@stoic dune

[stoic dune]

Mb, I think I got confused

None of these are fractions. They all have a denominator of 1

[peak crane]

yea

[stoic dune]

So the lcd would be 1, in that sense

[peak crane]

how

isnt it 1x

[stoic dune]

x doesn't divide 1

[peak crane]

what did u divide again?

[stoic dune]

The denominators. They are each 1.

[peak crane]

yea

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[lean otter]

How do i know whether the normal vector is directed upwards or downwards?

Lets say i want to find the shortest distance between point P_0 and the plane

I would wanna project the vector PP_0 onto the normal vector, which would give me a vector lets call it v, then the magnitude of v would be that distance but how do i know whether the normal is going upwards like i drew it or if its under the plane? And how can i know if the normal is shorter than the actual projection of PP_0?

Like that, how do i know if the normal isnt the blue vector? Then i wouldnt be able to project the vector PP_0 onto the normal no?

<@&286206848099549185>

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@lean otter Has your question been resolved?

[lean otter]

can someone help me?

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@lean otter Has your question been resolved?

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[lean otter]

For a question like this, I found sin x = 1/2 , -3

I disregarded -3 because that is out of the range for sin

therefore, I only used 1/2 to solve

[inner obsidian]

Should be right if all your math is correct

[lean otter]

After that, I noticed that since 1/2 is one of the solutions for sin, and since the value is positive, It means the answers are in quadrant one and 2

but

since sin 1/2 equals 30 degrees

and the reference angle is 30 degrees

we already found the quadrant one angle

but to find the quadrant two angle

do we do 90+30 degrees

or 180-30 degrees for the 2nd quadrant value

[lean otter]

this is the part i dont get

[inner obsidian]

uhhhh

i know exactly what you're talking about

i prob shoulda actually learned this stuff instead of memorizing

Sleep king here got you hopefully

Best of luck

[mental kettle]

Yes

[lean otter]

thats wrong tho

its 180-30

from the answer sheet

[inner obsidian]

oh yeah

wait yeah

[lean otter]

but why so

[inner obsidian]

because think about it

let me draw it out

[lean otter]

ok

[mental kettle]

Im bad at this

I tgink Because it will be 30 degree from the other enf

[mental kettle]

End*

[lean otter]

but why

[lean otter]

i dont get the drawing

[inner obsidian]

Yeah me neither but thats how i understood it

[lean otter]

lmao

[inner obsidian]

Here, my boy Ramonov got you

[lean otter]

alr lol

[thin bridge]

recall the unit circle definitions of trig functions

[mental kettle]

If we made a triangle it will have the same opp and hypo as the 30 degree

So the opp / hypo would br the same

Be*

[thin bridge]

and how reference angles are determined

you could also consider symmetry here

[inner obsidian]

yep, like my awesome drawing

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[safe radish]

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[desert mountain]

Can someone help me solve this question?

[desert mountain]

The goal is to solve for x

And I got x= -1.3

[vapid anvil]

Can you show your work

[desert mountain]

Alright

I used calcu for the last part probably why It seems off

[vapid anvil]

$4x + x = 5x$

[flat frigate]

ColdTee

[desert mountain]

Oh

Let redo it

-0.8

Did I do something wrong again

.close

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[rancid condor]

can someone please help me find aº and bº.

[inner lintel]

I don't see the solution yet. Are there any other clues? Instructions?

[rancid condor]

no

this is whats given

and the question is to find the pronumeral

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@rancid condor Has your question been resolved?

[spring glacier]

Circumscribed angle

[spring glacier]

You want this theorem

[safe radish]

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[terse remnant]

I am on part C of this problem, and the eigenvalues are 1, negative sq root of 5, and sq root of 5. Are the min/max just the highest/lowest eigenvalues? I don't quite get how to solve it if it is not

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[terse remnant]

.close

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[burnt stone]

can anyone help with this

[still charm]

Sure!!!

Make a u sub in the second integral

[hexed dune]

even function means that $\int_{a}^{b} = \int_{-b}^{-a}$

[flat frigate]

doctor99268

[burnt stone]

oh

did not know that

[hexed dune]

@burnt stone its self explanatory if you look at a graph

[burnt stone]

ok

[still charm]

Int_-4^-2 f(x) dx
Let u = -x, du = -dx
x = -4, u = 4
x = -2, u = 2
= int_(4)^(2) -f(-u) du
= int_(2)^(4) f(-u) du (flip bounds)
= int_2^4 f(u) du (even)

[hexed dune]

@burnt stone

[burnt stone]

ok

makes sense

ok now i understand how its C + D

Thanks

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[compact sierra]

not sure what i should be taking the limit off

[spring glacier]

n-> infinity

[compact sierra]

ya i know

i mean of what

(8+2n)^1/2 or(8-2)^1/2

or (8+2n)^1/2 -1

[spring glacier]

$\lim_{n \to\infty} a_{n}$

[flat frigate]

Cycadellic

[spring glacier]

Is what is wants

[compact sierra]

yes but what is an

that is my question

[spring glacier]

That sequence

[compact sierra]

lol brother get you give me more than that

[spring glacier]

We have a base term and a way to get the next term

Then we know a1

We can use the rule to find a2

Then a3 and so on

[compact sierra]

ya im still lost brother

[spring glacier]

Start with a1

[compact sierra]

ok the limit is -1

[spring glacier]

a1=-1

Then we work on to a2

$a_2=\sqrt{8+2a_1}$

[flat frigate]

Cycadellic

[compact sierra]

ok so rad 6

[spring glacier]

We keep doing that

a3, then a4, then a5

Keep going

Forever

What do we get to

[compact sierra]

thats it it just wants to plug and chug

[spring glacier]

No, we have to find a way to solve it, but thats what it means

Like

We want to find a function that does

$f(n)=a_n$

[flat frigate]

Cycadellic

[spring glacier]

A way to put n directly into the function so the limit is easier to take

Assuming we can

[compact sierra]

ok so rad 8+2n-1?\

[mellow trench]

I suppose you have to find a value of a_k for which the value of a_k+1 ceases thus ending this recursion

Put ak=4

[spring glacier]

Oh

Its an infinite nested root

[mellow trench]

See the recursion stops

[compact sierra]

man im still lost i just dont know what an should be for the function so i can take its limit

[spring glacier]

No this one is easier than that

Its an infinitely long nested root

[spring glacier]

We can keep plugging x into x however many times we want

Itll give us the sequence

If we solve x like this, it gives us the limit

[spring glacier]

What i mean is that if you went through and kept putting this sqrt(8+2x) into x, youd always have another x you can put it in

And so this is the infinite series

[compact sierra]

ok so I was supposes just do some algebra and solve for an?

[spring glacier]

Solve this

x if your limit

[compact sierra]

ok

[spring glacier]

Does it make sense of this is the limit?

[compact sierra]

ya somewhat

[spring glacier]

Okay

[compact sierra]

.close

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[spring glacier]

Good

[safe radish]

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[vapid anvil]

How do i simplify $x^{2log_{10}x - 3} = 100$

[flat frigate]

ColdTee

[quasi bison]

write $x$ as $10^{\log_{10}(x)}$ and eventually end up at a quadratic in $\log_{10}(x)$

[flat frigate]

Ann

[quasi bison]

alternatively, substitute $u := \log_{10}(x)$ right away and get the equation $(10^u)^{2u-3} = 100$, which also rewrites into a quadratic in $u$

[flat frigate]

Ann

[vapid anvil]

Im confused on the part where the x disappered, how did they do that?

From $x^{2log_{10}x - 3} = 100$ to $2log_{10}x - 3 = log_{x}10^2$

[flat frigate]

ColdTee

[vapid anvil]

So it becomes 2u^2 - 3u = 2

[vapid anvil]

But how did they do this?

[safe radish]

@vapid anvil Has your question been resolved?

[gaunt garnet]

The second expression is not $log_{10}x$ it's $log_x{10^2}$

[flat frigate]

Oğuzhan

[vapid anvil]

Yeah sorry i just noticed

[gaunt garnet]

Alright

[safe radish]

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[brisk grail]

Hello, do we use the same principles or steps on solving a 2x2 matrix using cramer's rule the way we would solve a 3x3 matrix?

[brisk grail]

like in a 3x3 theres 3 variables so we find dx,dy,dz and d

for 2x2 do only need dx,dy and d?

[quasi bison]

yes

cramer's rule works for any size of matrix

even tho it gets really complicated really quickly as n goes up

[brisk grail]

alright thanks

.close

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[obsidian vine]

Should the equation be <DBG = 45 or 90? (NUMBER 1 PART B)

[safe radish]

@obsidian vine Has your question been resolved?

[obsidian vine]

<@&286206848099549185>

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@obsidian vine Has your question been resolved?

[safe radish]

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[cosmic grove]

do you know the trig identity for cos(2x) ?

[silent roost]

No

i forgot .-.

[cosmic grove]

:/

cos(2x) = cos²x - sin²x

they gave you sin(x), you just need cos(x)

[split ether]

Actually finding cos(x) is unnecessary

Just use cos(2x) = 1 - 2sin²x

[cosmic grove]

thats one way too yea

[silent roost]

im confused

[silent roost]

sin is negative so like

I'd have more than one

[split ether]

Wdym

[cosmic grove]

???

[split ether]

There's just one solution

[silent roost]

ohh there's multiple??

[cosmic grove]

No...

[silent roost]

very confused

[split ether]

Like I said, there's just one

[cosmic grove]

sin(x) = -15/17
thats it

[split ether]

Just simplify 1 - 2 * (-15/17)^2

[silent roost]

isn't that sin negative tho

and result in an answer greater than one?

[long copper]

you're squaring sin x though

[silent roost]

oh shoot

that makes sense now

it goes away

I got an answer super duper close to one

[long copper]

how did you do it

Did you do sin -15/17 instead of just -15/17

[split ether]

,w 1 - 2 * (-15/17)^2

[flat frigate]

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Requested by A Lonely Bean#7998

[split ether]

Doesn't seem to be close to 1

[long copper]

if you do 1-2(sin -15/17)^2 then you get close to 1 so i think that's where they made a mistake

[safe radish]

@silent roost Has your question been resolved?

[silent roost]

Yeah it is

[long copper]

-15/17 = sin x and not x so you don't have to "sin it" again if you're putting it into that identity

[silent roost]

OHH

omg

yes

[long copper]

If you change cos (2theta) into 1 - 2sin^2(theta) it becomes easier to see the number of solutions

[safe radish]

@silent roost Has your question been resolved?

[silent roost]

Oh right

So it ends up being one

wait

so it's 0?

[long copper]

what did you end up with

like what equation

[silent roost]

1-2sin^2theta -sin^2 theta = 1

[long copper]

if you simplify that what do you get

wait nvm

misread the interval

[long copper]

This is correct, the only solution is theta = 0

[silent roost]

it's wrong 💀

[long copper]

it comes out as $\sin^2\theta = 0$ right

[flat frigate]

I can't believe you've done this

[long copper]

oh wait nvm im a complete idiot it is 2 solutions i forgot one more

i confused myself

sin theta = 0 at 0, 180 and 360

it should be 2

[long copper]

sorry i said it was correct i was wrong

[silent roost]

whoops

it's ok

thx

i also need help on this but ima be afk for a little while

[safe radish]

@silent roost Has your question been resolved?

[silent roost]

Student B did it wrong right?

[long copper]

yh i think so

[silent roost]

I'm not done but I think I did something wrong

Wrong thing

[long copper]

It should be $\frac{1 + \tan^2 \theta}{\tan \theta}$ at the beginning i think

[flat frigate]

I can't believe you've done this

[long copper]

If you do that then it will come out to 1 + tan^2 theta at the end

[silent roost]

ohhh

okok

also does this work out

[silent roost]

Are you sure it does?

[silent roost]

I was wrong this checks out

Which of the following is an equivalent expression to sin θ − (sin θ)(cos^2(θ))? It comes out to sin^3(theta) right?

[long copper]

$\frac{1 + \tan^2 \theta}{\tan \theta} \cdot \tan \theta = 1 + \tan^2 \theta \equiv \sec^2 \theta$

[flat frigate]

I can't believe you've done this

[silent roost]

they have that at step 2 don't they?

I'm trying to figure out how to simplify (sin 2θ)(csc θ)

[long copper]

use $\sin 2\theta = 2 \sin \theta \cos \theta$

[flat frigate]

I can't believe you've done this

[silent roost]

Ya ik

Wait I think I get it now

The sin thetas cancel

Leaving the 2 and cos

[long copper]

yes

[silent roost]

there's another one I have to simplify cosx over 1-sinx

ik it equals sec + tan but idk why

i just used a graph

[long copper]

you could try simplifying 1 - sinx over cos x then doing 1/that

[silent roost]

idk

there's this other simple one

but idk my identity

for this one

it's not sending

oh god

[long copper]

ive got 3 of the same picture lol

[silent roost]

well yeah it's that one ;-;

[silent roost]

that sounds like a lengthy way to do it

but ig lots of things cancel right

[long copper]

well i suppose so you only need the top and the bottom one

[silent roost]

does it end up turning to 0?

[silent roost]

how about

[gaunt garnet]

You don't have to use those

[silent roost]

ye i was talking ab the first the actual thing u use

or the work i meant sorry

like

cos pi - theta
cos pi cos theta + sin pi sin theta
-cos theta + 0

sin pi/2 + theta
sin pi/2 cos theta + cos pi/2 sintheta

cos theta + 0

[gaunt garnet]

First find sin(theta) then go up to sin(2theta) by drawing the special right triangle

I will send the special right triangle I'm talking about right now

[gaunt garnet]

This should do the trick

Or just use sin(2x) = 2sinxcosx

[silent roost]

oh interesting

[gaunt garnet]

This is probably a better choice in this case*

[silent roost]

if I plug in cosx how do i find sin

[gaunt garnet]

Draw a triangle in the unit circle

But in the quadrant IV

[silent roost]

ohh ok i was confused that makes sense

13 is the hypotenuse

[gaunt garnet]

Yes

[silent roost]

how'd u find 12?

Pythagorean theorem?

[long copper]

yes

[gaunt garnet]

Yes

Or you can memorise it

3-4-5
5-12-13
7-24-25

[silent roost]

so sinx is 12/13?

[gaunt garnet]

Yes

2 * 12 / 13 * 5 / 13

[silent roost]

so then plug it all in

yep

120/169

[long copper]

it's in the 4th quadrant

[silent roost]

negative oops

I'm still trying to figure out where he made the mistake here ik u said in the start but his start makes sense to me

I thought step 2 was the problem

[long copper]

i'd say it's the step 2 moving into step 3 that's the problem

[silent roost]

yeah it shouldn't be squared right it should be just tan

1+ tan on the numerator

[long copper]

no

[silent roost]

im confused on that part

[gaunt garnet]

If I haven't remembered something wrong, this should be the table

^ means change name, sin -> cos, cos -> sin, tan -> cot

[silent roost]

ohh right yeah i think that's right

[gaunt garnet]

You can memorise this table

[silent roost]

i solved it using sum and difference identities but that works too

[gaunt garnet]

And there is actually a pattern

First third second, first third second (the plus signs)

You just have to memorise the left part which isn't that hard either