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[eager iron]

In this differential equation How does the y(p) become x^2/2-x?

If we set y=Ax+B we’re one degree to low so we need to set
y=Ax^2+Bx+C
y’=Ax+B
y’’=A

Replace the y’’ and y’ = x in the equation and we get

A + Ax + B = x

So (A)x = 1
(A+B) = 0

If A is A B must be -1

go back up to our y=Ax^2+Bx+C and plug in our values for A and B and we get

y = x^2 - x

so where is the /2 coming from?😭😭

Nvm i didn’t derive

Xd

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[candid vault]

Hello so how do you get the measure of CAG

[languid hinge]

14º is for what angle? FCG or FCE?

if it's for FCE, we can figure out the angle DAF by calculating 180-44-14

because G is the incenter, dividing the result by 2 would give us the angle CAG

[lone heart]

@candid vault Has your question been resolved?

[candid vault]

FCG I assumed

[candid vault]

I ended up doing same just with 44 and 28 lol

Thanks

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[sly nacelle]

ok nevermind i figured it out

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[alpine sable]

uh..

[alpine sable]

Im just geniunly confused

Nevermind I figured it out

[tawny condor]

If you don't have anything else to ask, do .close

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@alpine sable Has your question been resolved?

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[gritty pond]

how can i show A~A?

[tacit arch]

use the definition

and find the invertible matrix

[gritty pond]

so i think we WTS A = P^-1AP right?

[tacit arch]

yea find P

[gritty pond]

i switched it *

but basically

swap B with A?

im still confused cuz i dont remember much matrix

[tacit arch]

Use A = A

[gritty pond]

how is that supposed to help?

i already got the answer, i just dont get why it is

[tacit arch]

Yea that's P

P = I_n

[gritty pond]

i thought thats the identity matrix

[tacit arch]

Find the inverse of P

it is

[gritty pond]

why is P the identity matrix

Oh

is it cuz it sdays EXISTS

so only one suffice?

I still dont get

Why A = I_nAI_N

is identity matrix basically like 1 in basic algebra?

when it comesto matrix multiplication?

[tacit arch]

yes. do the multiplication to convince yourself

n=2 or n=3

[gritty pond]

Is there any case in which the identity matrix doesn't act as 1 like basic algebra?

[gritty pond]

for second solution i dont get what "multiplying both sides on the left/onthe right" mean

algebraically isn't it just "multiply both sides":

[worn fox]

Matrix mult isn't commutative, multiplying on the left and on the right are different in general

[tacit arch]

if you have never multiplyied matrices before, i suggest you do a few problems

[gritty pond]

ohh i think i get it

so say if we want to show B = P^-1AP => A = PBP^-1 we can do this?

so it looks like we have a new decision in algebra when its matrix multiply both sides of eq, left or right

i wonder if we can also do it from the middle?

can any1 confirm this is true???

and for this one, can i multiply both sides but from the middle?

or is it only the left/right?

[naive valley]

no, the right hand side should be A = P^-1 B P

[gritty pond]

this is new problem

the P-1's are switched

[naive valley]

you get that from the left hand side by left-multiplying by P^-1 and right-multiplying by P

[gritty pond]

it is an implication symbol

i WTS A~B => B~A

[gritty pond]

basically its same as this one but switch P-1 and P

[gritty pond]

this is my real problem

So disregard everything else to remove confusion

[naive valley]

ok

so what's your question about that

[gritty pond]

i used the online source

to basically copy the same thing

wut u think

for reflection

[naive valley]

yep, that is fine

P = I does the job

[gritty pond]

now for symmetric

i think this is what we want?

cuz its just flipping the thing

the vars*

or am i mistaken?

that arrow thingy is the implication syumbol with a question mark on it

[naive valley]

if B = PAP^-1, then A = P^-1 B P, not PBP^-1

[gritty pond]

how is that so

it makes sense when u matrix multiply it

but this is what we want to show,

unless we do something else compared to just flipping the variables?

[naive valley]

start with B = PAP^-1
left-multiply both sides by P^-1 to get:
P^-1 B = AP^-1
now right-multiply both sides by P to get:
P^-1 B P = A

[gritty pond]

yes i said that mazkes sense

but for my WTS it is different

[naive valley]

ok, so what you wrote is incorrect

[gritty pond]

because i think we have to flip the variables

or how do we find what we want to show

for symmetry

[naive valley]

i would simply say:

[gritty pond]

because thats what i always thought we had to do for symmetry, flip the variables

[naive valley]

now let Q = P^-1

then Q plays the desired role

A = QBQ^-1

[gritty pond]

OHHH wait

I think i know

Is it because the equivalence relation is saying "there exists an invertible matrix P "

so it doesnt have to be the same?

[naive valley]

yes correct

it's not the same P when you are showing B~A

but that's fine, you just need an invertible matrix

the P that works for B~A is the inverse of the P that works for A~B

that's why i called one of them Q above, to avoid confusion

[foggy pecan]

If you become familiar with notation, you can write this:

$A\sim\text{}B\Leftrightarrow B=PAP^{-1}\Leftrightarrow\\A=P^{-1}BP\Leftrightarrow A=P^{-1}B\left( P^{-1} \right)^{-1}\Leftrightarrow B\sim\text{}A$

[ocean seal]

Joanna Angel

[naive valley]

yes perfect

[gritty pond]

hm

for some reason i always thought it had to be like P P^-1 to cancel out

not also P^-1 P

[naive valley]

nope, inverses work on both sides

for rectangular, non-square matrices, there's such a thing as a left inverse that is not a right inverse, or vice verssa

but for square matrices, if a matrix is a left inverse, then it's also a right inverse, and vice versa

ah you shouldn't start by assuming it's the same P in the first two equations

i would start with:
B = PAP^-1
C = QBQ^-1

since P and Q may not be the same matrix

then you're looking for an invertible matrix, say R

such that C = RAR^-1

[lone heart]

@gritty pond Has your question been resolved?

[gritty pond]

oh so thats still simple i think?

so we just do C = PQAP^-1Q^-1 and let R = PQ?

[naive valley]

should be QPAP^-1Q^-1 ,not PQAP^-1Q^-1, but that's the right idea

R = QP

yes exactly

(may want to explicitly comment that (QP)^-1 = P^-1 Q^-1 if that's not an already well established fact at that point in your material)

[lone heart]

@gritty pond Has your question been resolved?

[lone heart]

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[undone saffron]

Idk how to do this proof without working on the other side

I am apparently only supposed to work on one side

[unkempt mica]

expand csc and you are pretty much done

[undone saffron]

okay let me try

and I can cancel out the sins

and I am left with 2/cos^2x?

which is 2sec^2x?

[unkempt mica]

yes

[undone saffron]

ohhh thanks

[unkempt mica]

nw

[undone saffron]

also

you know how like sin(90-x) = cos

is that also the case with sin(x-90)

like is sin(x-90)=cos?

[unkempt mica]

no but almost

[undone saffron]

wait can you explain

if it does not equal that how would I find sin(x-90)?

[unkempt mica]

do you know that sin(-x) = -sin(x) ?

[undone saffron]

yeah

but not for cos or sec

because cos(-x) = cos(x) right

does this have anything to do with it

[unkempt mica]

sin(x-90) = sin(-(90-x)) = -sin(90-x) = -cos(x)

[undone saffron]

oh

I don’t know what to do at the end again

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@undone saffron Has your question been resolved?

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[tame dragon]

is 8 and 16 like terms

[ocean whale]

Does 8 and 16 have the same variable and power?

[tame dragon]

its just 8 and 16

no power or varible

[ocean whale]

It technically does

x^0 = 1

[tame dragon]

rightt

[ocean whale]

So both can be written as 8x^0 and 16x^0

So do they have the same power and variable?

[tame dragon]

yes

[ocean whale]

So then are they like terms?

[tame dragon]

yes

[ocean whale]

There you go

[tame dragon]

tyy

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[grizzled burrow]

eh bisects feg, yes r no

[lone heart]

Please don't occupy multiple help channels.

[grizzled burrow]

i think so

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[grizzled burrow]

.reopen

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✅

[naive crystal]

yeah it bisects

[grizzled burrow]

but why

[naive crystal]

FEGH is cyclic meaning you can draw a circle passing through F,E,G,H

[grizzled burrow]

explain please

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@grizzled burrow Has your question been resolved?

[grizzled burrow]

.close

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[golden blaze]

Are you able to check my work on this assignment please

[lone heart]

@golden blaze Has your question been resolved?

[lavish rapids]

Sure, one sec

[lavish rapids]

For the first question, you should consider what the absolute value means before putting a negative within the brackets. The second question has the right idea and correctly shows the reflection but it may be helpful to plot some key points first since some of the points seem inaccurate, for example, is the square root of 5 equal 4?

[golden blaze]

So for number 1 I should put the - outside the brackets?

But for 2 where would I plot the points at?

[lavish rapids]

Yep! Put the - outside the brackets to ensure all numbers are made negative

[lavish rapids]

Think about some of the perfect squares, things like 1, 4, 9, 16, etc.

They will provide a nice number to work with and a nice outline

[lavish rapids]

So for example, when x is 4, the y will be 2, when x is 9, the y will be 3

[golden blaze]

This work?

[lavish rapids]

Yep! That’s much more accurate

[golden blaze]

Quick question for number 1 I knew it was reflect but not sure if it was x or y, if it reflected across the x axis would the opening face the left or right?

[lavish rapids]

If it was reflected in the y nothing would change, since it’s symmetrical along the y axis

Wait I’m confused, what exactly do you mean?

[golden blaze]

It’s fine I get what I did wrong

[lavish rapids]

Oh, alrighty

Other than the first two, most of them seem right, but there’s few that could probably be tweaked or corrected

For this question, I would make a slight adjustment to the formula for the first graph, and then for the quadratic I would maybe fix the key points a little bit too

For the first graph it might help to think what you’d have to multiply root 2 by to get 2

[golden blaze]

So 2x in the square root?

[lavish rapids]

Along with this one, you may want to check over the key points of the second question

[lavish rapids]

Yep! That’s right

[golden blaze]

Where would I plot the point on the 2x^2 question?

[lavish rapids]

Try going through it by finding out what the y is when you sub in x

So when x = 1 and -1 looks right, but what would y be when x = 2 or -2?

[golden blaze]

Would it just double as well? So I would plot my points at (-2,4) and (2,4)?

[lavish rapids]

If the equation is y = 2x^2

Then what is 2 x 2^2

[golden blaze]

So (-2,8) and (2,8)

[lavish rapids]

Yep!

Is there anything I did that you’d want to go over again? Or anything else you don’t really understand?

[golden blaze]

On the most recent picture u sent and the bottom question Would the transformations be reflects across the x axis and translation?

[lavish rapids]

Yep, everything was right, the only problem was that you forgot to apply the a value

[golden blaze]

This work?

[lavish rapids]

Yep! You’ve got it

[golden blaze]

Alright I think that’s it for now, thanks a bunch for your help

[lavish rapids]

Np

Have a good day or night

[golden blaze]

U too

[lone heart]

@golden blaze Has your question been resolved?

[lone heart]

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[alpine sable]

seems right to me

what you said

[vale wigeon]

f(0) undefined tho

f(x) = 4/x

f(0) is undefined

yes but (x-3)/(4/x) simplifies to (x^2 - 3x)/4 only for x ≠ 0

[alpine sable]

yes but youve already simplified it

i think its like a hole

[vale wigeon]

for x = 0 the expression (x-3)/(4/x) is still undefined

yes it is a hole exactly

[alpine sable]

like if you have (x^2-9)/(x-3)

becomes x+3

[vale wigeon]

parentheses

[alpine sable]

but hole at 3

math site did a funny i guess

[vale wigeon]

it has a hole at 0 yes

[alpine sable]

ye

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[quaint light]

i cant understand what a set of real.valued functions on an interval 8s

is

is it all the real valued funtions that map an element in the interval to another element in the interval?

[echo socket]

All functions with domain [0, 1] and codomain R

[hushed locust]

by "real-valued" they mean outputs a real number, so in this case it means functions that map the interval S to the real numbers

[quaint light]

@hushed locust @echo socket

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[cinder breach]

can someody please explain the steps?

[cinder breach]

hi

yes sure

[ruby current]

try multiplying the first fraction by m/m and the second by n/n

[cinder breach]

thanks

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[gray patio]

Honestly I am so lost with the n=k+1 step

I have no idea what to sub in

cause there is no like equation on the L.H.S

[fresh coral]

ok, based on your own formulations, what is P_1?

[gray patio]

P_1 is 0.6

[fresh coral]

ok, and P_2?

[gray patio]

How would you calculate that

that may help actually

cause there are two cases

[fresh coral]

are you familiar with making like a probablity tree

[gray patio]

You are right

let me try that

13/25

p_2

But say for the n= k+1 term how are u meant to

make a tree?

[fresh coral]

start with a tree

oh

im silly

you should use induction here, obviously, as the question states

[gray patio]

NO thats why i was confuised

ahahah

but

[fresh coral]

ok, so what is the inductive step here?

[gray patio]

it made me like understand what was going on more

Im stuck on the n= k+1

i did base case

and assumed n= k

but how can i prove n= k+1

[fresh coral]

you just gotta expand it out

[gray patio]

expand what?

[fresh coral]

what is the expression of P_{k+1}

[gray patio]

just p(n)_ but with 5^k+1

[fresh coral]

yeah

what is another way to write 5^k+1

[gray patio]

5 x 5^k+1

but from there ?

am i missing smth??

[fresh coral]

so what's another way to express P_{k+1}, if we have P_{k}? use the information that there's a 60% chance if it rained the previous day and 40% if it didn't

[gray patio]

is there not 2 cases

so itll be

o.6 x p(k) x 0.4 x p(k)?

[fresh coral]

0.6*P_k + 0.4 * (1-P_k)

[gray patio]

1-P(k)???

I understand the first bit

[fresh coral]

the probability of it not raining

is 1-x

[gray patio]

OML

You are so right

i did not clock that

whatsover

thank you

sm

You are life saver : )

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[burnt escarp]

7^x + 2 = 5^x + 1

[burnt escarp]

how would I solve this quesiton using logs

[quiet vector]

7^(x + 2) = 5^(x + 1)?

[frail grove]

is it $7^x + 2 = 5^x + 1$ or $7^{x + 2} = 5^{x + 1}$

left or right ?

[ocean seal]

Adam Chebil

[burnt escarp]

right

the second one

[velvet sapphire]

take logs both sides

[burnt escarp]

wdym?

[velvet sapphire]

$(x+2)\log{7}=(x+1)\log{5}$

[ocean seal]

阿皮pea

[burnt escarp]

oh

[velvet sapphire]

$\log{7^{x+2}}=\log{5^{x+1}}$

[ocean seal]

阿皮pea

[velvet sapphire]

do this first

and i guess u are able to solve it

[burnt escarp]

wait

I am a bit confused

is the base for the log 7?

or is it 10

still

[velvet sapphire]

10

[burnt escarp]

oh

so they have the same base

can I not just drop the logs?

[velvet sapphire]

yes

[velvet sapphire]

wdym by dropping the logs?

[burnt escarp]

if we have a log on each side of the equal sign with the same base we can drop it right? to equate for x

[velvet sapphire]

so after u drop it, it will become $x+2=x+1$?

[ocean seal]

阿皮pea

[velvet sapphire]

definitely not

[quiet vector]

from here?

look at the base of the exponents

7 and 5

[burnt escarp]

yeah

[quiet vector]

why would you equate them

they're clearly not equal

u still stuck?

[burnt escarp]

yeah...

[velvet sapphire]

u can do this after u added the logs

[burnt escarp]

ok wait

[burnt escarp]

so after did this

how do we continue

or do I move around the logs?

[velvet sapphire]

both log7 and log5 are constants

[quiet vector]

yea

so treat them like numbers

just expand

[velvet sapphire]

yeah

[quiet vector]

also

[burnt escarp]

oh

[quiet vector]

.

[burnt escarp]

your so right

I got it

[quiet vector]

@velvet sapphire

[burnt escarp]

I expand

[quiet vector]

bs w me @velvet sapphire

[velvet sapphire]

what

[burnt escarp]

how about

[quiet vector]

[burnt escarp]

B

4. B

Previous years test btw

[quiet vector]

lets start with a simpler question

[burnt escarp]

I did the rest above

just both question of 4

seem very tricky ngl

[quiet vector]

yea

thats why a simple one first wait

[burnt escarp]

ok

[quiet vector]

$log_2 x = 5$

[ocean seal]

Shah2044UwU

[quiet vector]

sorry im trash with latex so

now can u slove this

@burnt escarp

[burnt escarp]

yeah

[quiet vector]

so whats the solution

[burnt escarp]

wait

[quiet vector]

||our target is this tho||

[burnt escarp]

wait why is this confusing me

no not htat

log x / log 2 = 5

[burnt escarp]

Wait

how do I solve this again

[quiet vector]

dang uve quite a few formulas at ur disposal dont ya

[burnt escarp]

yeah

[quiet vector]

only the log definition needed here tho

[burnt escarp]

ye

[quiet vector]

so when i put a number in log base 2

im asking 2 raised to the power what equals that number

x in this case

[burnt escarp]

yeah

[quiet vector]

and the answer seems to be 5

soooo

[burnt escarp]

so

log 5 / log 2

right?

[quiet vector]

the value of x...

its not log 5/log2

[burnt escarp]

oh.

uk what ill ask another time it seems I am a bit too tired to understand this

[lone heart]

@burnt escarp Has your question been resolved?

[lone heart]

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[charred sedge]

whats the y coord of the vertex in 2(x-3)(x-2)

[lethal belfry]

!status

[lone heart]

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

[raw stream]

y intercept?

you mean

[gray isle]

same idea as before

[charred sedge]

3

i got -1/2 but i went on a website and i was told i was wrong

[lethal belfry]

if you know calculus, find the turning point using derivatives

[charred sedge]

ok thanks

[gray isle]

show exactly what the website asked for
and what you entered

[raw stream]

I misread the question sorry

[charred sedge]

i went on chat gbt and i asked it what is the vertex of 2(x-3)(x-2) and it said (5/2, -25/2)

[gray isle]

...

!nogpt

[lone heart]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[charred sedge]

so would it be (5/2,-1/2) then?

[gray isle]

yes

[raw stream]

If you are in doubt always graph it in desmos

[charred sedge]

what website do you recommend to check answers

[lethal belfry]

or use wolfram alpha

[fallen verge]

Not chat gpt

[hearty nest]

yes

[raw stream]

Well I would use that as last resort

try solving it on your own

[charred sedge]

ok

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[clear comet]

How do I write if x could be any number?

[alpine sable]

u mean $x \in \mathbb{R}$
?

[ocean seal]

Snöwdinger

swagmafia3000

[clear comet]

like the equation returns x=x

[cinder tundra]

He said any number, not any real number tho

He must specify first

[ocean seal]

swagmafia3000

[alpine sable]

Like in sets or in calc?

[clear comet]

calc

[clear comet]

this could be right

[cinder tundra]

If your equation returned this you made a mistake

[clear comet]

no

[ocean seal]

swagmafia3000

[cinder tundra]

!original

[lone heart]

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.

[clear comet]

in the solution key it also says any number is possible for x

[cinder tundra]

Non related to my statement

[clear comet]

xeN is good enough for me

its just a homework

[cinder tundra]

That’s why the context is important

[clear comet]

thanks everyone

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[royal kiln]

Hi, trying to understand this solution

[royal kiln]

https://i.imgur.com/OPluWPR.png

https://i.imgur.com/lHoYywa.png

I'm stuck on the "Note that..." after the tan alpha/beta

my geometry is failing me

I don't get how angle CPD = angle CQD + QCP + QDP

I also thought angle QCP = angle QDP....

[prime river]

In triangle APB angle P is exterior angle for triangle APD so that equals <A+ <D
<A = <Q+<C bcz it is exterior angle for triangle ACQ

[lone heart]

@royal kiln Has your question been resolved?

[prime river]

<cpd=<apb vertically opposite angle

<apb=<pad + <pda exterior angle property

[royal kiln]

sry I was away, just didn't wanna close it before I get it

i think i gotta draw this out

I totally forgot exterior angle property is even a thing, and tbh I'm still not getting it but I think I can look that up now

[prime river]

<cpd=<apb vertically opposite angle ....1
<apb=<pad + <pda exterior angle property ... 2
<pad= <cqd + <qca exterior angle property
Using in 2
<apb =<cqd+<qca+<pda

[royal kiln]

I don't see this <pad= <cqd + <qca exterior angle property

... I was reading that line wrong and was very confused

ok makes sense now lol

thanks

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[bright scarab]

!status

[lone heart]

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

[bright scarab]

4

i think that y value shouldnt reach negative

then the parabola should have same roots

so D = 0

[zealous lichen]

having a root means the equation = 0 at the root

but we want > 0

[lone heart]

@bright scarab Has your question been resolved?

[lone heart]

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[blazing junco]

Hi, here is my thoughts. I know trigonometry and even further, but idk why 5th and 4.1 is actually wrong

[bright scarab]

bro

[blazing junco]

hi

[bright scarab]

wrong

[blazing junco]

are you talking about 4th or 5th?

I didn't get it

<@&286206848099549185>

[bright scarab]

1st

[blazing junco]

can you explain pls?

[lone heart]

@blazing junco Has your question been resolved?

[blazing junco]

.close

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[alpine sable]

i don't get the concept of pi and pi radian like how pi radian = 180degree and only pi = 22/7, like i get why pi = 22/7... it's a value, but i don't quite understand pi 'radian'

[ocean seal]

! What the hell am I doing here?

[alpine sable]

sorry aprox

[worn fox]

https://tenor.com/view/radians-math-angles-circle-pi-gif-20830003

[slender gull]

Why would you wait so long for that gif lol. Could have saved some words from me.

[alpine sable]

makes much sense now...

i could be wrong, correct me if i got it wrong...

[worn fox]

lool sorry came in here after your message. its a neat gif, first one when you search "radian" (dont look at the rest)

[alpine sable]

ohhhhhhhhhhh

okokokokokoo

it took 3 times

like 3 cycles

and the last part were the decimal digits

did i got it right?

[worn fox]

sure

you need pi many radians to go halfway around the circle

[alpine sable]

wdym

@worn fox

[worn fox]

im not sure what you dont understand

in the gif, 3 rads isnt enough to go around half the circle

you need pi radians

[alpine sable]

oh ok

so basically if we use the length of the radius as the the measurement device to measure the circumference of the circle and name that unit as rad, so after 3 rad it's not sufficient to cover up the half of the circle so we use the pi decimal digits to cover till there...

did i got it right?

if yes

then i am good then

i don't have any doubts...

[slender gull]

That's what it is, yeah. A radian is the angle an arc of length = radius subtends.

[alpine sable]

hmmm

thank you for teaching me in the simplest language possible... :)

thank you @slender gull @worn fox

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[austere charm]

Ok so when I use soh cah or toa

Do I use the one where I don't know either answer

Like since I know the hyp so I use toa?

[wise spruce]

Huh

You use the one with a side you know and the side you wanna find

You wanna find the opposite and you have the hyp so you use soh (o/h)

[austere charm]

Ok so tan 18.9 degree then what blank/76cm

[wise spruce]

Why are you using tan

[austere charm]

Sorry sin*

But what is the math

All I know is sin18.9

[wise spruce]

What does sin represent

[austere charm]

The angle?

Idk

[wise spruce]

What does soh cah toa mean

S = … C = … T = …

[austere charm]

Dude I know that

I just don't know if I times

The other stuff

While dividing

[wise spruce]

Well sin = opposite / hypotenus right

So if you take the sin of the angle

Which is opposite / hypotenus

And multiply it by the hypotenuse

The / and x cancel

To get the opposite side

[austere charm]

So what is like the equation

Written out

So I can understand better

[wise spruce]

Sin 18.9 degrees (o/h)

x 76 (xh)

To get o

[austere charm]

So after I sin the degrees or what ever I do in the question I times it by the one side that I allready know?

[wise spruce]

Yes

[austere charm]

So 64.62

24*

24.62

[wise spruce]

Yes

[austere charm]

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[austere charm]

ty

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[south yarrow]

Hi there, I just need a sanity check, been years since I've done these and trying to help out someone else. The question asks to find the value of k, where k makes up 2 real roots of the equation 4x^2 - 8x + k = 0. I've only been able to find one root, however, and keep coming back to the idea that the question may be incorrect, and 2 real roots don't exist?

[south yarrow]

Not thought about these in a solid 5 years

[mortal trellis]

the question doesnt really make sense

do you mean to find k such that k itself is one of the roots of the polynomial?

[prime river]

Two real roots when b²-4ac ≥ 0

[mortal trellis]

well if =0 then one real root. with multiplicity 2

[south yarrow]

Hang on, I'll send a pic of the Q

The question seemed poorly worded/designed to me

[mortal trellis]

so you want that the roots x_1 and x_2 are equal

which by the previous discussion means that the discriminant should be 0

[south yarrow]

Yeah, I thought exactly the same thing, but the question made me think there we're 2 differing roots, so I doubted my math

Thanks for the help

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[royal kiln]

Can someone help me understand the first part of this solution? https://i.imgur.com/LCHU8VI.png

[royal kiln]

https://i.imgur.com/yupadRd.png

I don't see how that first equation follows from geometric progression of the tan of angles

I'm reading geometric progression as
tan(EAC)
tan(EAD) = n tan(EAC)
tan(EAB) = n^2 tan(EAC)

[lone heart]

@royal kiln Has your question been resolved?

[royal kiln]

I also (though this might be the same issue) just don't get how

tan^2(EAD) is equal to tan(EAC)tan(EAB)

someone help me understand this 😭

[royal kiln]

I think I got it... I didn't realize the order they gave the angles in was the order they were in in their respective geometric/arithmetic progression

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[unkempt mica]

have you tried anything?

[weak wharf]

Just do implicit differentiation

and solve for dy/dx

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[craggy forge]

Find the quadratic equation, in standard form, that has the following root:
3-sqrt6 and 3+sqrt6

[craggy forge]

do we add the roots and divide by two to get x vertex value, but I do not remember how to proceed

[pale bone]

P(x)=(x-x1)(x-x2) where x1 and x2 are the roots that you mentioned

[craggy forge]

how do I do it if my roots are two terms

Since I can’t combine 3+sqrt6 or 3-sqrt6

Oh

Do I make it sqrt54

and -sqrt54

[prime river]

If a and b are roots then equation is
k ( x²-(a+b)x+ab)=0 where k constant

you can take k=1

a+b= 3+√6+3-√6=?

[craggy forge]

x^2-6x+3

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[hexed eagle]

i need to find where this wave has constructive and destructive interference

[hexed eagle]

so i took the derivative and set to 0

nvm

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[tropic yacht]

I’m lost like a mofo

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[waxen flame]

What are you confused about?

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[soft isle]

How do i find all real zeroes ?!?!?

[soft isle]

when i used synthetic division and got zero i tried to find the rest of the zeros but i couldn’t factor the polynomial i got

the 50x^3 + 35x^2 - 20x + 5.. what do i do after i got that?

[hidden fable]

I think your division is wrong

try doing it again

[soft isle]

the -2/5 was a given zero and when i divided it i got 0 and i don’t know what to do after that

[hidden fable]

I know, but your result in the division is not right

[soft isle]

wait what do you mean?

[hidden fable]

it's 50x^3 + 35x^2 - 20x - 5

[soft isle]

i accidentally changed -5 to +5

[hidden fable]

yeah

[soft isle]

so what comes after that?

i’m still confused

[hidden fable]

so yeah, my first step would of course be to try to factorise it, as you said

[soft isle]

ok i’ll try

i factored it and got this?

wait can i not use factor by grouping?

do i have to factor the entire thing by 5?

[hidden fable]

ok a good step would be to divide everything by 5, it will make the numbers a bit wasier to work with

[soft isle]

ok i’ll try

[hidden fable]

you may also notice something

[hidden fable]

you can try a few simple values

[soft isle]

what do you mean?

[hidden fable]

you can try a few values of x that make sense in this case, just to see if we get any luck

[soft isle]

as in plug in values of x in the equation?

[hidden fable]

mhm

if you take a close look, there is a pretty easy value that can make that zero

and once we have that, finding the other two is a lot simpler

[soft isle]

i’m sorry i really don’t know the value, i can’t think straight

i tried putting in 4

[lusty hearth]

rational root theorem would help

[soft isle]

ooh

1/2 is another zero??

[lusty hearth]

yep

[soft isle]

now do i do synthetic division using 1/2? to find the rest?

[lusty hearth]

yep

[soft isle]

wait should i have just used the possible rational zeros from the previous result of -2/5 instead of doing synthetic division again

[lusty hearth]

wdym

[soft isle]

when i used the rational root theorem and got the possible rational zeros of the function

the function i divided by 5

[lusty hearth]

i recommend dividing first since if you get it to a quadratic, you can potentially factor

which is quicker than rational root theorem

i gtg sorry

[soft isle]

ok thanks for your help

i did what you said and i still can’t factor

wait nevermind i get it

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[copper wadi]

I just started back with Calc II after a break from Calc I. A bit rusty with integration. Could someone give me pointers as to where I’m going wrong?

[limpid spade]

Where did -5 come from

[stable pecan]

When you expand the second equation back, you will realise where you made the mistake

[copper wadi]

Oh lol

I see

Thanks!

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[ebon meadow]

so as of now, i have my proof, i just want to be very precise with my variables but not sure how to be

this is what i have

i want to add more to "a, b, c are elements of R"

[harsh swallow]

why?

[ebon meadow]

it says that we need to be extremely precise with our explanations

[harsh swallow]

This isn't a difficult proof

as in

[ebon meadow]

for example, a, b, c all can't equal each other, or else this proof would be false

[harsh swallow]

it doesn't require very many assumptions and/or subproofs

you take a, b, and c to be arbitrary elements of R

which is what $a, b, c \in \mathbb{R}$ means

[ocean seal]

Katharine

[ebon meadow]

i know, but we have to mention certain things like this

i'm just curious if there are some other specific things that i need to mention

[exotic canopy]

i mean

$a, b, c \in \bR$ means "take 3 arbitrary real numbers"

[ocean seal]

artemetra

[ebon meadow]

so 3 different ones?

[harsh swallow]

arbitrary

meaning any value in R is allowed

[exotic canopy]

can be different, can be same – the point is that it works for both

[harsh swallow]

no other limitations apply

for example you could say

[hidden fable]

you can also provide a counterexample yeah
if it's false for one set of values, then it's already a false statement

[vale crag]

just go give a counter example if you're that scared of your examiners

[harsh swallow]

$a, b, c \in \mathbb{R} | a > 0$

[ocean seal]

Katharine

[harsh swallow]

in which case a has to be bigger than 0

but b and c are any real number

counter example like they said

or just state, since lhs $\neq$ rhs if a $\neq$ c

[ocean seal]

Katharine

[harsh swallow]

something like that

[ebon meadow]

okok

thanks

.close

[lone heart]

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[real cedar]

hello?

[harsh swallow]

hello

[hidden fable]

hi

[real cedar]

i need help with some questions but first are these correct?

[hidden fable]

sorry for asking, but, what's this from?

[real cedar]

i have a pre-test

im rlly confused tbh

[hidden fable]

are they like practice problems?

[real cedar]

yes

[queen jay]

i assume so

[hidden fable]

okay so the first one is okay

and in the second one you had the right idea, but take a look at the direction of the < and >

[real cedar]

wym? like tbh they kinda slapped me ina face witht this i went for easy practice to this tbh idk

[lone heart]

@real cedar Has your question been resolved?

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[vapid shuttle]

No

sqrt(x) returns only positive values

sqrt(9) is only 3

sqrt(25) is only 5

What did I say that makes you think I’m doing code??

No worries use a different word if you want

The square root function, outputs/returns/gives/makes only positive values

The confusion comes from comparing an equation like

$$x^2=25$$
To
$$x=\sqrt{25}$$

[ocean seal]

Austin

[vapid shuttle]

Because they are different equations. Completely

If you wanted to solve the top equation

The answer would be x=5 AND x=-5

Right?

(-5)^2=25

But…. If you want to solve the bottom equation, the sqrt as a function only gives positive values, the answer to the bottom is ONLY x=5

$$\sqrt{x^2} \neq x$$ $$\sqrt{x^2}=|x|$$

[ocean seal]

Austin

[vapid shuttle]

Yes and the rule is stated above ⬆️

[ocean seal]

David L. Bowman

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[golden elbow]

Find the derivative:

[golden elbow]

cos(x)tan^2(x) - 2tan(x)sec^2(x)sin(x) / tan^4(x)

cos(x)tan(x) - 2sec^2(x)sin(x) / tan^3(x)

the 2 possible answers are

I did the formula correct and it's correct but I don't know how to translate to different forms and it looks like thats a skill they want out of us

[pseudo ice]

what are you trying here?

[golden elbow]

I am trying to do a derivative

of sinx / tan^2x

[pseudo ice]

Oh is that equals supposed to be a - sign?

[golden elbow]

that = meant to be a -

yes thanks

[pseudo ice]

Figured

One thing that might help is to note that sec^2(x) = sec(x) * sec(x) then remembering how sec is defined, and playing around with that a bit

[golden elbow]

they only have one option for g and one option for f though so maybe we can choose any of them?

sec = 1 / cos

[pseudo ice]

That might be a point

[golden elbow]

so cos(x)tan(x) - 2 (1/cos)(1/cos)sin(x) / tan^3(x)

[tardy stag]

lol the fact that only one of the functions uses t as a variable, gee i wonder which one its derivative is

[golden elbow]

right now i'm focused on v(x)

I don't want to solve just because the letters match

I want to learn the calculus

[pseudo ice]

[and I don't think so, think this one is a trick one where you need to be careful (esp noting the options for u, I think they want you to check signs!)]

[golden elbow]

but just because the signs are whatever, doesnt mean I solved the calculus

that's using your eyes to get around the calculus

[pseudo ice]

Yep, looking good, now that second term of the numerator

[golden elbow]

cos(x)tan(x) - 2sin(x) / cos^2(x) / tan^3(x)

[pseudo ice]

there's a particular reason I suggested to write it as sec * sec, and that wasn't to combine them (I mean, you can, but there's a better way!)

[golden elbow]

cos(x)tan(x) - 2sin(x) sec*sec / tan^3(x)

cos(x)tan(x) - 2sin(x) (1/cos) (1/cos) / tan^3(x)

like that?

so tan^3(x) = sin^3(x) / cos^3(x)

so if we flip it it takes away the fraction cos and just becomes cos^2(x) with no denominator anymore

oh actually there would be sin

the weird thing is I got a B+ in Calc 1 at my university

shows how bad this university is

this is Calc 1 review

[pseudo ice]

Was thinking that only one of the sec need to be 1/cos, but in any case in the end you'd need to change it so I guess

[golden elbow]

can you tell me some more hints

this is not for marks or anything

I'm just trying to learn

cos(x)tan(x) - 2sin(x) sec(x) 1/cos(x) / tan^3(x)

cos(x)tan(x) - 2tan(x)sec(x) / tan^3(x)

[pseudo ice]

maybe think about what sin(x) * 1/cos(x) is

[golden elbow]

yea I did that

[pseudo ice]

Yep there it is One extra factor of tan(x) in the numerator for you

[golden elbow]

ok so then it can become

cos(x) - 2sec(x) / tan ^2(x)

if my math is this weak and it's week 1 of Calculus 2 and this is Calculus 1 review and it's been 10 years since I did Calculus 1 should I just drop out and practice?

[pseudo ice]

actually I don't think it's any of the options

[golden elbow]

am I doomed for Calculus 2 if my Calculus 1 is this weak?

[tardy stag]

it is one of the options

[golden elbow]

I need honesty though

am I doomed for Calculus 2 if its started this week?

I've had to drop it like 3 -4 times in the past

ok

so cos(x) - 2sec(x) / tan^2(x) right?

[pseudo ice]

yea it is, I confused myself with what I had written on my scribble paper

[golden elbow]

ok chart can you give me a hint please?

[pseudo ice]

Anyways from here, there are different ways you could go about it, you could for example then try and convert tan^2(x) into sin^2(x)/cos^2(x) and work with that, or maybe think about the numerator...

[golden elbow]

yea I just did that

so I get cos^3(x) / sin^2(x) - 2cos(x) / sin^2(x)

[pseudo ice]

Well sin^2(x) as the first denominator but yea

[golden elbow]

yea

ok good

so I got that

factor out cos(x) ?

[pseudo ice]

You could might be a good idea to also factor out a 1/sin(x) too at the same time(!)

[golden elbow]

cos(x)/sin(x) ( cos^2(x) / sin(x) - 2/sin(x) )

that look good to you?

[pseudo ice]

this is getting a bit more messy than I had thought

But that should work I think first one should be + though right?

[golden elbow]

yea

cot(x) ( cos(x) cot(x) - 2/sin(x) )

I think I'm going to cry

[pseudo ice]

[I will say that simplifying at the start makes your life a lot easier!]

But maybe instead convert that cos^2 into something else...

[golden elbow]

1 - sin^2(x) ?

[pseudo ice]

Yep in here, then life should become a bit closer to what we want(!)

[golden elbow]

cos(x) / sin(x) (1 -sin^2(x) / sin(x) - 2 /sin(x) )

[pseudo ice]

Yep, and you can simplify the stuff inside the

[golden elbow]

cos(x) / sin(x) (sin(x) - sin^3(x) - 2 )

cos(x) - cos(x)(sin^2(x)) - 2cos(x)/sin(x)

[pseudo ice]

gonna type this out so it's nicer to see
[
\frac{\cos(x)}{\sin(x)} \pqty{\frac{1 - \sin^2(x)}{\sin(x)} - \frac2{\sin(x)} }
]

[ocean seal]

@pseudo ice

[golden elbow]

oh the top is just cos there

cos^2(x)

[pseudo ice]

Seeing those terms already have a common denominator to begin with...
[
\frac{\cos(x)}{\sin(x)} \pqty{\frac{1 - \sin^2(x) - 2}{\sin(x)} }
]

[golden elbow]

cos(x) / sin(x) ( cos^2(x) / sin(x) - 2cos^2 + 2sin(x) / sin(x) )

[ocean seal]

@pseudo ice

[pseudo ice]

Then try to play around with that maybe for a while

[golden elbow]

cos(x) / sin(x) ( cos^2(x) - 2 / sin(x) )

cos(x) / sin(x) (2sin^2(x) - cos^2(x) / sin(x) )

[pseudo ice]

don't convert back to cos (as that's what we just came from, that's going backwards )

[golden elbow]

cos(x) / sin(x) ( 1 - sin^2(x) - 2 / sin(x) )

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

but you wan't me to stay away from cos?

cos(x) / sin(x) ( 1 / sin(x) - sin(x) - 2 / sin(x) )

[pseudo ice]

That is true-

[golden elbow]

cos(x) / sin(x) ( -sin(x) - 1/sin(x) )

[pseudo ice]

we used that here

[golden elbow]

-cos(x) - cos(x) / sin^2(x)

[pseudo ice]

first one is -cos(x)

But otherwise yep, then try and convert that second term!

[golden elbow]

yea the equal sign

always hit that instead of -

-cos(x) - cos(x) / sin^2(x)

-cos(x) - cot(x) 1/sin(x)

do you think I should drop this class and rethink my life?

[pseudo ice]

Yep, and then there's one more change you can do to that and you have one of your options(!)

[golden elbow]

ohhhh 1/cos = sec and 1/sin = csc

-cos(x) - cot(x)csc(x)

you are so amazing

[pseudo ice]

There you go

[golden elbow]

thank you

so just some honesty

am I doomed for Calculus 2?

[pseudo ice]

Anyways, also, if you simplified at the start, you could have written sin(x)/tan^2(x) = cos(x) * cot(x) then that one should hopefully be easier to differentiate