help-33

[ornate coral]

its not supposed to be 2y its supposed to be 1y im pretty sure..

[chrome mirage]

the 4th line isn't right

[muted narwhal]

its 2+y-1-y right

[late geode]

can't drop those bases like that

[chrome mirage]

no actauy

[ornate coral]

just take 2 power 2+y as 2power2 x 2powery and similar with 2 power 1+y and take 2 power y common

[chrome mirage]

a^(m+n) = a^(2)
= > m+n = 2

a^m - a^n = a^2
not equal to m-n = 2

[muted narwhal]

ah

thought so

well I'm struggling to combine those together, if they can be combined at all

[chrome mirage]

2^(2+y) = 2^2 * 2^y

[muted narwhal]

huh you can do that

I was thinking about doing something along those lines

lemme try again then

where did the n come from

theres no n in my question

[humble mist]

im just asking bro

[muted narwhal]

oh, you need to open a new help for that

[humble mist]

How ?

Oh bruh thanQs

[muted narwhal]

could you delete your stuff then pls

[humble mist]

Yes of course

[muted narwhal]

thanks

[humble mist]

wlcm

[muted narwhal]

@chrome mirage I'm kind of stuck

I seperated everything so now I got x^2 * 2^y - 2^1 * 2^y

what now?

can i merge the 2^y together

[chrome mirage]

u can either take 2^y as common or take it as another variable

[muted narwhal]

ahh

[chrome mirage]

like letting 2^y = a

that'd be easier

[muted narwhal]

I see

so now I have 2^(2)a - 2a = 2^4

factorise 2 out?

and the a of course

[chrome mirage]

4a -2a = 2^4

normal sum no need to take anything or factorrize

[muted narwhal]

4a -2a =2a

where did the other a come from?

for the RHS

[chrome mirage]

u have 2^4 in right side

[muted narwhal]

mhm

[chrome mirage]

oh my bad

yea 2^4 in RHS

[muted narwhal]

oh so its 2a=16

[chrome mirage]

yep

[muted narwhal]

oh I see now

y=3?

[chrome mirage]

yes

[muted narwhal]

perfect

thanks for the help flamingo

[chrome mirage]

[muted narwhal]

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[marsh citrus]

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[vague rivet]

Hello, im in stats and I need help on jotting certain things down

[vague rivet]

I did the math and all but im confused by some of the questions, ill upload them rn]

questions: The average SAT verbal score at private schools is 450 (with a standard deviation of 100). A new
private school promoted itself by advertising that its graduates had higher SAT verbal scores. At the
end of the school year, the Parent–Teacher Association (PTA) decided to test whether graduating
seniors obtained higher SAT verbal scores than average. The average score was 466 for 83 graduating
seniors at the new private school. (Use alpha = .01)

What is the test statistic (z score) for the sample on the comparison distribution?
b) What is the p-value?
c) Do you reject or fail to reject the null hypothesis?
a. *** And explain why based on NHST Method.
b. *** And explain why based on P-value Method.
d) What is your conclusion about the significance of the results?
e) What is your conclusion about the research hypothesis?

last questions before my answers: (z cutoff is 2.33) Confidence limit method (label the z cutoff, the confidence limits- lower and upper,
the percentage of the confidence interval, and where the population mean falls)

[hard creek]

ap stats type shit

[vague rivet]

im in college stats lol

you got it

last questions before my answers: (z cutoff is 2.33) Confidence limit method (label the z cutoff, the confidence limits- lower and upper,
the percentage of the confidence interval, and where the population mean falls)

thats the part that has me confused

[hard creek]

I am in calc tbh I dont have the slightest idea

[vague rivet]

1. for a one tailed study isnt there only an upper confidence limit?

[hard creek]

my knowledge goes to standard deviation and mean absoloute deviation

[vague rivet]

ahh

do yk about confidence limits and stuff?

2. i dont know how to jot the population mean properly

It’s “3)”

[marsh citrus]

@vague rivet Has your question been resolved?

[marsh citrus]

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[wind mauve]

Been stuck on getting the point for point slope form. I inputted a few different answers but I don't understand how any of the information would get me a point dependent on t.

[runic temple]

What is x equal to at time t?

[wind mauve]

is it cos(t)?

[runic temple]

Yes

[wind mauve]

i assume sin(t) for the other one then

[runic temple]

Yes

[wind mauve]

thank u

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[frail lodge]

can someone please explain how to they took the partial deriative?

did they first expand this out?

or did they just apply product rule as is?

in particular, any fast ways of computing this?

[marsh citrus]

@frail lodge Has your question been resolved?

[marsh citrus]

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[lyric kelp]

have to solve for tetra

[marsh peak]

Recall that cot = 1/tan

[lyric kelp]

yes i put it in 1/tan

[marsh peak]

You can turn that into a quadratic equation in tan

[lyric kelp]

OH

so i can use

quadratic formula?

with sin cos tan?

[marsh peak]

Yes

[lyric kelp]

i didnt know it was allowed haha ill try

uhm is this alright

what i did

[viscid sky]

Yea

[lyric kelp]

i got 3 and 1

this feels really cursed

[viscid sky]

thats just tan theta

not theta

[lyric kelp]

so tan tetha =3

can i find the angle with tan-1

[viscid sky]

we know for 1 is tan 45

for other we will write it in inverse form

[lyric kelp]

ah its 71.57

is it safe to say

since tan is positive that

it will be in the first and 3rd quadrant?

[viscid sky]

idk depends on question

but usually we mention in inverse form

[lyric kelp]

question was Solve for θ, giving exact solutions where possible, 0◦ ≤ θ < 360◦
.

oklay its good

tysm!

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[stray token]

Got a bit of a dumb question here: is a^2^k the same thing as a^2k ?

[cunning basin]

if you mean $(a^2)^k$ then yes

[elfin berry]

WhereWolf(ping if needed)

[cunning basin]

if you mean $a^{(2^k)}$ then no

[elfin berry]

WhereWolf(ping if needed)

[stray token]

Hmmm

I thought they were the same

I guess that would be the error

Thanks for the help

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[proud crag]

HElp

[proud crag]

is my answer correct

please anyone

[mental panther]

yes

[proud crag]

is it correct?

[mental panther]

yes

[proud crag]

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[marsh citrus]

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[jovial shadow]

9. Sam deposited P5,000 in a corporation
   that earns annual interest rate
   compounded quarterly. His money grew to
   P5520 in 1 year. What is the annual
   interest rate?

[still temple]

do you know the formula for compounded interest rate

[jovial shadow]

Yep wait

P - 5,000
F - 5520
M - 4
T - 1
Is this correct?

[still temple]

yesh

yeah

just calculate r now

[jovial shadow]

Okay thank you!

It's 0.100173739x100 right?

So the final answer is 10.02?

[still temple]

you can check if you want to

try to get F using the r you just got

if it turns out to be 5520, then you are correct with your calculations

[jovial shadow]

It workedd! But 10.02% must be in decimal which is 0.100200000

[still temple]

very weird but yeah I guess that works haha

[jovial shadow]

Thank you so much! Can you also review my other answers?

[still temple]

to the best of my abilities sure

[still temple]

that's a lot 😅

[jovial shadow]

Yeah🥲 but dw it's just identification i didn't include the solving

[still temple]

is general annuity like

compounded?

sorry idk terms Really

oh guess not

welp sorry I don't remember those terms tbh

[marsh citrus]

@jovial shadow Has your question been resolved?

[marsh citrus]

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[vestal tree]

how did delta x become 1

[vestal tree]

one the right side it becomes -1 on top and 1 on the bottom

[vernal forge]

1+delta x is 1

as delta x goes to 0

[vestal tree]

delta x is also 1

[vernal forge]

no

it approaches 0

therefore 1+delta x approaches 1

[vestal tree]

then how did these two lines happen

sorry im really confused

[vernal forge]

cancel out delta x on the numerator and denominator

[vestal tree]

ohhhh

ohh okay

okay i get it now

thank you

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[fierce thunder]

x^2 − 2x + 10 = O(x^2)

[fierce thunder]

how do i prove this without using lim

[marsh peak]

Well can you recall the definition of O?

[jovial shadow]

Yep like simple annuity is that the compounsing period and payment period are the same but in general they are not

[marsh citrus]

@fierce thunder Has your question been resolved?

[fierce thunder]

the thing is i cant get my head around the theory about it

like i get the upper bound lower bound

and n_0 is like "where" an expression can or cannot overtake a function

[fierce thunder]

you have any resources to understand this?

still cant get it after 3 hours lmao

[marsh peak]

No I don't really have any resources

But you can show that x^2 - 2x + 10 < 2x^2 for x > sqrt(11) - 1

[fierce thunder]

2x^2 for x > sqrt(11) - 1 how did you get this part?

[marsh peak]

Solved x^2 - 2x + 10 < 2x^2

[fierce thunder]

where did 2x^2 come from?

[marsh peak]

I chose the coefficient to be 2

[fierce thunder]

ah is that the cg(x) something like that

[marsh peak]

Because, clearly, 2x^2 will outgrow x^2 - 2x + 10

Yes

[fierce thunder]

wait lemme try to solve it

wait how do i solve x^2-2x+10<2x^2, like what do i isolate

[marsh peak]

x^2 - 2x + 10 < 2x^2
10 < x^2 + 2x
11 < x^2 + 2x + 1
11 < (x + 1)^2

And out of sqrt(11) - 1 < x or x > -sqrt(11) - 1 we would obviousny want the former

[fierce thunder]

whats the purpose of adding 1?

is that like what im supposed to do at all problems or?

[marsh peak]

It's to complete the square

[fierce thunder]

okay i got the formula

but why do i need to get that value

sorry i dont get like the concept of what were doing lol

from what im understanding if the value is sqrt(11) - 1 that means x^2 -2x+10 is less than the other one right

[marsh peak]

We found out that if x > sqrt(11) - 1, then x^2 - 2x + 10 < 2x^2

[fierce thunder]

but how does that relate to the O(x^2)

[marsh peak]

Yes

[fierce thunder]

OHH

WAIT

ohh

i get it now lmao

it finally clicked

[marsh peak]

And don't forget that |x^2 - 2x + 10| = x^2 - 2x + 10 since it's always positive

[fierce thunder]

got it thank you again!

now i got the concept wtf

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[fierce thunder]

.reopen

[marsh citrus]

✅

[fierce thunder]

@marsh peak hello sorry for pinging but what if its -x^2 +5x + 1 < ?

can my right hand side be -2x^2?

or should it just be 2x^2

[marsh peak]

Depends on what you are trying to show

... = O(x^2) again?

[fierce thunder]

yes

the difference is only the -

[marsh peak]

Well first you need to take the absolute value of -x^2 + 5x + 1

[fierce thunder]

ohh i see

so the right hand side would be 2x^2 right

since its bigger

[marsh peak]

Yeah you can pick 2x^2

[fierce thunder]

got it thank you again!

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[flat junco]

can somebody guide me how to proceed further?

[marsh citrus]

@flat junco Has your question been resolved?

[marsh citrus]

@flat junco Has your question been resolved?

[marsh citrus]

@flat junco Has your question been resolved?

[marsh citrus]

@flat junco Has your question been resolved?

[marsh citrus]

@flat junco Has your question been resolved?

[west lotus]

hi

Is this a correct triangle?

[marsh citrus]

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

[marsh citrus]

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[real locust]

I'm having a comical amount of trouble with this

[real locust]

I'm going through soh coa toah videos but the it's being explained is just confusing me

[jaunty creek]

,rotate

[real locust]

That's a nifty feature

[jaunty creek]

You need to use cosine for this triangle:

Cos(30) = a/8

[real locust]

Okay gotcha

[jaunty creek]

👍

[real locust]

Thank you

[jaunty creek]

Np

Good luck

[real locust]

Siryessir

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[marsh citrus]

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[real locust]

Wait

How do you open up the channel again for other people

[marsh citrus]

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[real locust]

Back again

[real locust]

I'm following along with what I'm watching but it's with a problem that has a number in A's place

Only thing I can work with here I think is 30 and 8

[earnest siren]

I’m having a bit of trouble understanding. What is the issue again?

Is it a different problem?

[real locust]

Oh I'm sorry let me grab the question

[real locust]

This is the initial problem

[earnest siren]

Ok

[real locust]

I haven't taken any math classes in a hot minute so forgive me for not being good with any of this

[hollow sparrow]

no worries

do you know SOHCAHTOA?

[real locust]

I've been looking into it for the past 30 minutes or so but the way it's being explained is confusing me

Mostly because they're most if not all doing problems with 3 numbers or two numbers to fill in stuff

[hollow sparrow]

no worries let's walk through this

[real locust]

I got a angle and a 8

[hollow sparrow]

so trig functions are ratios of sides of a right triangle

and we have just defined them as such:

[real locust]

Right

[hollow sparrow]

an easy way to remember this is to just remember SOHCAHTOA

[earnest siren]

Just a question. Are you having trouble applying sin and stuff

Or understanding how it works

[real locust]

I'd say yes

Hard yes

[earnest siren]

So which one?

Or both?

[hollow sparrow]

so looking at your problem, the angle is at the top. a is close to it and we have the hypotenuse. So which trig function should we use? Is a opposite or adjacent to the angle?

[real locust]

As I understand it we'd use cos

I get tripped up pretty often when the shape gets rotated

It's just from me forgetting and overthinking if flipping it around changes anything important

[earnest siren]

Never seen someone use the logical or

[real locust]

Logical or?

[earnest siren]

Never mind

[real locust]

Alrighty

[earnest siren]

So what equation do you get

[real locust]

I got as far as I posted above

[earnest siren]

So if you multiply by 8 you get the value of a right?

In the first equation

[real locust]

I think??

Hold on

I'm going off this

It's screwing me up at where 4 + 7 is

Cause by what I've figured that's where A would go

Don't know what to do with A

[earnest siren]

I don’t think you need to do anything after find a

[real locust]

So I was right?

[earnest siren]

8*cos(30) = a

I think that was all you needed

The khan academy one was doing something different

In your problem you are given h for free while in khan academy they used Pythagoras to find it

[real locust]

I see

[real locust]

This will sound dumb but is that the answer or another step?

[earnest siren]

Depends

Do you know what cos(30) is

[real locust]

I don't off the top of my head

[earnest siren]

If you need in radical form

[real locust]

My calculator says

[earnest siren]

Cos(30)=sqrt(3)/2

If you need it rounded use the calculator value

[real locust]

0.8660254038

Is what I got

[earnest siren]

For cos?

Multiply by it by 8

And round

[real locust]

Okay gotcha

[earnest siren]

Or

[earnest siren]

Use this

And leave the radical

[real locust]

6.982

There's alot more numbers but I didn't wanna type them

[earnest siren]

Yeah it rounds to that anyway

[real locust]

Ok gotcha

Can I ask a favor?

[earnest siren]

Sure

[real locust]

If you're able or want to, could you type out how you did all this

?

[earnest siren]

Maybe

[real locust]

Im Translating this to notes but I have trouble writing the stuff sometimes since I don't 100% understand all of it

Haven't touched math in 2 years

[earnest siren]

Ok one moment

Cos = adjacent / hypotenuse

From SOH CAH TOA

Cos(30) = a / 8

It has to be over 8 as it is the longest line and so it is the hypotenuse

The adjacent side is a

And the opposite side doesn’t have a name

8 * cos(30) = a algebra

You can plug this into a calculator

Or use the radical form

Cos(30) = sqrt(3)/2

One moment

If you sub it in for cos(30) and simply you got 4sqrt(3)

This is the same as what you get when you round in a calculator straight from 8cos(30)

[real locust]

Okay gotcja

Thank you

[earnest siren]

No problem

[real locust]

Would you be down to help with a few more problems or you good?

[earnest siren]

I can’t right now, maybe in 20 mins

[real locust]

Bet

@hollow sparrow real quick, what trig operation would I use here?

[floral linden]

you have adjacent and want to find opposite

which operation connects the 2

[real locust]

Cos

Rigjt

It's cos

LETS GOOOO

[hollow sparrow]

See? It just takes a few practice times and you’re good to go

[real locust]

Let's go baby I'm learning math

[marsh citrus]

@real locust Has your question been resolved?

[real locust]

https://tenor.com/view/no-rat-nuh-uh-nope-no-no-gif-11699607390227676176

[marsh citrus]

@real locust Has your question been resolved?

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[green wind]

I'm confused on the wording of this question... what exactly is it asking for?

[marsh citrus]

@green wind Has your question been resolved?

[green wind]

<@&286206848099549185> Take your time. The server required me to ping you. I am working on other problems in the meantime.

[hasty ruin]

derivative by first principles
find the original function and at which point the derivative is being evaluated at

[green wind]

so would the answer for 'f' on part a be 1? how could i figure out what the a value would be for part a?

[hasty ruin]

you can write out the formula for differentiation by first principles and basically substitute everything in

[green wind]

is the formula for differentiation by first principles the same as the difference quotient? which is as i understand it: f(x+h) - f(x) / h

.close

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[ashen crown]

Can someone help me with this?

[ashen crown]

I dont understand where the 9 and 15 comes from

I got it!

.close

[marsh citrus]

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[vague fog]

Need Help

[somber sapphire]

help me with q 1

[hard creek]

okay

A

[somber sapphire]

.close

[marsh citrus]

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[vague fog]

.reopen

.close

[hard creek]

whats up Tami

[vague fog]

I need help

[vague fog]

I did not close the room

@hard creek

[hard creek]

yeah

do you need help with this

[vague fog]

I need help with the one's above as well

[hard creek]

there are a ton there just pick 3 Ill explain them then you can try to do the rest on your own

alright

for this one

since the slopes and the y intercepts are consistent throughout all the answer options we can immediatley conclude that the question is asking about which ineqaulity sign is correct

so we must observe the graph

since the shaded area is below the line it must be less than or less than / eqaul to

since the line is dashed it must be less than

hence A is the correct answer

[vague fog]

Got it.

I really need help with these nine for the test I have next week

I know it's a lot

[vague fog]

Got that

[hard creek]

aight lets go to the next one then

okay send the next one

[hard creek]

so what is a negative slope

in your mind

[marsh citrus]

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[hard creek]

Ill tell you right now B's slope is underfined and C's slope is 0

[vague fog]

a negative slope is a line that is trending downward from left to right

[hard creek]

so what is wrong with B

[vague fog]

B's slope is underfined and C's slope is 0 like you said

[hard creek]

is B moving from left to right or just up to down

[hard creek]

exactly which leaves A and D

which out of the two is trending downwards from left to right

[vague fog]

A

[hard creek]

okasy next one

[vague fog]

Is it A??

[hard creek]

yeah

[hard creek]

aigfht for this one we need to use the slope eqaution

y1-y2/x1-x2

do you know this

[vague fog]

Half/half

[hard creek]

aight Ill explain

(x1,y1) represent one point and (x2,yx) represent another point

to find the slope between them we must calculate the change in y over the change in x

hence

(y1-y2)/(x1-x2)

do you follow?

[vague fog]

Yes

[hard creek]

aight

plug the values in

it should give you 2

I belive

lemme know what you get

[vague fog]

It woud be 3/2 since we are doing it in simplest form

[hard creek]

what no it should be 2

my bad

silly error

6/3=2

[vague fog]

My bad, same. Cat is on my lap

[hard creek]

(5--1)/(3-0)=(6)/(3)=2

lmao

[vague fog]

On my keybaord and everthing lol

[hard creek]

aight

same idea

[vague fog]

Sorry bad pic

[hard creek]

yeah

send a better one rq

[hard creek]

aight

so find delta y / delta x

okay this one should be -3/2

[vague fog]

I was about to type the same thing

[hard creek]

should be B

lets reason about it

the y-intercept has to be -4

the graph has to have a downward trend from left to right

only B meets that criteria

[vague fog]

I follow

Ohhhh'

[hard creek]

okay to the next one

hmm

you do this one

y-intercept is -2

the graph has to have a upward trend from left to right

only one such graph

[vague fog]

Hmmm

One sec

[hard creek]

it is B

[vague fog]

I got (0, -2), (1, 2), and (-1, -6).

For the points so, yeah it aligns with B

[hard creek]

next one

as you know I am in a rush to go to sleep

[hard creek]

oh god

[vague fog]

I know

[hard creek]

should be a

lets examine it step by step

[vague fog]

Ok

[hard creek]

we know there is an upwards trend from left to right

right?

now we can eliminate B and D

then potentially we can plug and chug using the point (4,3) and see which one satisfies that value

or

we can approximate slope to be less than 1 hence it must be A

either way

reaches the right answer

[vague fog]

Right'

[hard creek]

okay

do you understand this one

[vague fog]

3 left

[hard creek]

if so let us go to the next one

[hard creek]

same idea

is the tren from left ot right upwards or downwards?

upwards

[vague fog]

Upwards

[hard creek]

now we can eliminate C and D

[vague fog]

Sorry, mine was delayed

[hard creek]

now is the slope greater than 1 or less than 1

as i mentioned this stratedgy doesnt always work as both options wont always be on opposing sides of 1 but here it is fine

in this case the slope is greater than 1 hence we can conclude the answer is B

[vague fog]

2 left

[hard creek]

now if we wanted to do this properly we could find delta y/delta x

next one

[hard creek]

okay step one

consider the function itself

y-intercept of 1

upward treng left to right

lmao

no further math neccesary

[vague fog]

Lmao

[hard creek]

A B and C are all obviously wrong

bc their y-intercepts are wrong

however let us further confirm by now examining the ineqauylity

it says less then

meaning the line must be dashed and shaded below

is that true for D

yes it is

[vague fog]

It is.

??

[hard creek]

true

lets go to the next one

[vague fog]

Last one

[hard creek]

okay

no workarounds here

except...

for the immediate conclusions from the slope

negative slope vs positive slope

2x has a positive slopes so C is wrong

since we said a slope of 2 and d has a slope of 3 d must be wrong

[vague fog]

So is d

[hard creek]

no

[vague fog]

So it's A or B

[hard creek]

now we must consider A and B

now greater than or eqaul to means the the line is filled and is shaded above

hence.......

drum roll please

the answer must be B

https://tenor.com/view/dance-gif-14880344851904561392

[vague fog]

Thank you

[hard creek]

no worries

[vague fog]

Sleep well.

[hard creek]

do you understand at least?

[vague fog]

A lot more

[hard creek]

good

dont always use the workarounds I showed you for mcs

[vague fog]

Better than when My teacher explained it. I won't

[hard creek]

but they are always helpful in time crunches as they enable you to eliminate options

[vague fog]

True

[hard creek]

they are very hel;pful for SAT and ACT

[vague fog]

For sure

[hard creek]

but now always great to use on free response

anywaysbest of luck with you work

[vague fog]

Thank you

.close

[marsh citrus]

@hard creek Has your question been resolved?

[marsh citrus]

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[muted narwhal]

can someone quickly answer me this, we have function sqrt(4-x^2) and I need to take the inverse, I'm currently on x^(2)-4 = -y^2

[muted narwhal]

so do I bring the - or the square to the other side first

[devout mauve]

to be able to "bring the square to the other side" you have to take sqrts

which you cannot do if one of the sides is negative

which -y^2 is

[muted narwhal]

oh I forgot, "bring to other side" is not the proper definition

yea do I multiply -1 or sqrt both sides first

so I got to multiply both by -1 then

[devout mauve]

like I said, you cannot do sqrt on both sides right now

[still temple]

$$y = \sqrt{4-x^2}$$
$$-y^2 = x^2-4$$

[elfin berry]

<:F_button:1095679234497843251>

[devout mauve]

also note that 4-x^2 is positive, so x^2-4 is negative

so by multiplying by -1 you make both sides positive

[muted narwhal]

I see

so it is ok for a unknown number like x to be negative

[devout mauve]

why shouldnt it

[muted narwhal]

square roots cannot take negative numbers?

but x isn't really any number

yea I guess that makes sense

[devout mauve]

4-x^2 cant be negative, thats correct

for y=sqrt(4-x^2) to be defined

but x definitely can. eg x=-1 works perfectly fine

[muted narwhal]

ah theres a limit for x >/= 2 yes?

[still temple]

UUUUUUH

If the task is to just find the inverse, you just have to solve for x

[muted narwhal]

oh I switched x and y around and rearrange to make x the subject

[devout mauve]

not sure what >/= is supposed to mean

[still temple]

yeah...

[devout mauve]

-2 <= x <=2

[muted narwhal]

i mean y

sorry :V

ok I'm getting a bit ahead of myself for this question

[still temple]

Did you find the inverse yet?

You're not... D is kinda over complicating it

[muted narwhal]

I think so?

[still temple]

You don't have to worry about the domain at this point

[muted narwhal]

sqrt(-x^2+4)

[still temple]

$$-y^2 = x^2-4$$
You're at this step

[elfin berry]

<:F_button:1095679234497843251>

[still temple]

so isolate x^2 by putting it own side of the equation

then figure out how to turn it into just x

[devout mauve]

you should always worry about whether stuff you write down actually makes sense

[still temple]

yeah, they can worry about it after they write it down

bro has not even written down teh inverse yet and you're talking about it's domain already >_<

[devout mauve]

i am talking about the domain of the original function

[still temple]

why tho

[muted narwhal]

the domain of the original function is -2 to 2

[devout mauve]

and they asked what x can be

[muted narwhal]

hmmm

I got y=sqrt(4-x^(2))

but answer page says -sqrt(4-x^2)

[devout mauve]

thats wrong

[muted narwhal]

the answer is wrong or I'm wrong

[devout mauve]

the answer cannot have constant sign

I just noticed the function doesnt even have an inverse

!original

[marsh citrus]

Please show the original problem, exactly as it was stated to you. A picture or screenshot is best.

If the original problem is not in English, then post it anyway! The additional context might still help helpers help you. Do your best to translate.

[muted narwhal]

this is becoming more complicated than I wanted it to be

[devout mauve]

ok so the domain is -2<=x<0

that's important

on that domain the function does have an inverse

lets look at the step before the sqrt

what do you have

[muted narwhal]

oh thats a question

x^2-4=-y^2

[devout mauve]

after multiplying by -1

[muted narwhal]

-x^(2)+4=y^2

[devout mauve]

or y^2=4-x^2. so now we want to take sqrt on both sides

we then have to decide whether we want the positive root or the negative root

yes?

[muted narwhal]

does that determine which side were using i.e the negative or positive side on a graph

and since domain is -2, 0, therefore we take -?

[devout mauve]

yes

[muted narwhal]

ahhh

[devout mauve]

we want the value that we get to be in our domain again

so it has to be negative

[muted narwhal]

so thats why theres a negative outside

[devout mauve]

yes

[muted narwhal]

ahh that makes much more sense

then the domains restricted to 0 and 2.....????

actually nvm I got it

thanks for the help denascite

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[marsh citrus]

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[fierce thunder]

how do i simplify 2nlog(n) + (nlogn)^2?

[devout mauve]

you cant simplify it that much

[fierce thunder]

wait but it is possible right?

[devout mauve]

well define possible

and what kind of result you want

[fierce thunder]

whats the simplest form can it do but its still log

no ln or e or something like that

[devout mauve]

because ln and log are so different

[fierce thunder]

this is what he got but im not sure how to

[devout mauve]

well the last term is (ab)^2=a^2b^2

where a is n and b is logn

[fierce thunder]

so i just distribute the exponent to both n and logn?

[devout mauve]

yes

[fierce thunder]

oh im dumb got it

thank you

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[marsh citrus]

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[harsh vine]

how do we calculate the volume of this if we are only given 1 bound

[floral linden]

x^2 - 3x intersects the x-axis at 2 points

im assuming the bounds are those points

[harsh vine]

ah shit yeah

at 0 and 3 im assuming

[floral linden]

yes

[harsh vine]

alright thanks

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[marsh citrus]

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[fierce thunder]

x^3+3x-1 < 2x^3

[fierce thunder]

how do i simplify this

[vernal forge]

subtract x^3 from both sides

[fierce thunder]

x^3-1<x^3

[marsh peak]

Are you still on the big O?

[fierce thunder]

yep

still practicing problems

[vernal forge]

it's 3x-1<x^3

[fierce thunder]

but now its just simplifying

[marsh peak]

There's an elegant proof to show that any polynomial function is big O of the leading term though

No need to struggle over every polynomial

[fierce thunder]

:?

[fierce thunder]

so it would be cube root of 3x-1 < x ?

[marsh peak]

$\abs{a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0}$, start by applying triangle inequality on this

[elfin berry]

A Lonely Bean

[fierce thunder]

ngl i dont understand anything in the picture haha

ithink its a pattern though

[marsh peak]

I took absolute value of an arbitrary polynomial there

[fierce thunder]

i dont think i could learn that test is tom :

[marsh peak]

Will you be asked to prove something like x^3 + 3x - 1 = O(x^3)?

[fierce thunder]

yep

[marsh peak]

I mean with this one proof you will just be able to give the answer immediately

[fierce thunder]

mmm

[marsh citrus]

@fierce thunder Has your question been resolved?

[marsh peak]

So, are you familiar with triangle inequality?

[fierce thunder]

nope not a clue

[marsh peak]

$\abs{x + y} \le \abs{x} + \abs{y}$, never seen this?

[elfin berry]

A Lonely Bean

[fierce thunder]

my prof did something like this

[fierce thunder]

i dont think so

absolute value i somehow know where u just turn the - into positive

[marsh peak]

Yeah it looks like they picked n >= 1 here

[fierce thunder]

whats the difference between the method you said and this pic?

[marsh peak]

Here they chose the value of n_0 and found out how large c needed to be

In my method I did it the other way

I chose the value of c and found out how large n_0 needed to be

[fierce thunder]

so what you would do is 5n^2 + 150n + 750<= 6n^2 right?

[marsh peak]

Something like that, yes, but I later found it that it may be better to pick n_0 = 1

[fierce thunder]

so doing the method in the picture?

[marsh peak]

Do you want me to explain what they did here?

[marsh peak]

Yes

[fierce thunder]

i think i get if i think hard enough haha

from what i understand

when its at n_0 = 1

you need a constant c

which c determines what you want

so make the cg(x) bigger when n_0 = 1 to prove that the left hand side is smaller than the right

is that somehow right?

[marsh peak]

Yes

[fierce thunder]

ohh kinda get it

wait but in proving it

do i still need to substitute c in the orig equation or just computing for c is fine?

i have to right because thats the proof or am i misundertanding?

[marsh peak]

No, it's done when you have found c I think

But I would still ask the professor how they want it to be solved

[fierce thunder]

got it, ithink ikinda get the concept now

but my algebra is braindead haha forget all the fundamentals 💀

[marsh citrus]

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[subtle imp]

What are the steps for finding a from A,b and c. Using the cosine rule.

[sleek nimbus]

the cosine rule is using only one angle, I think you mean sine rule

[subtle imp]

from A,b and c

[sleek nimbus]

ohh nvm

you can just Put that into the equation

[subtle imp]

I got sqrt(4*cos(145)) = c

[sleek nimbus]

send in you're work

[subtle imp]

Q1)c.

sry for handwriting

[sleek nimbus]

if you need to find a and you have b c and A then a needs to be in the right side of the equation

[subtle imp]

don't worry about the letters themselves, just their use

[still temple]

I was asking can we not just right sin 145 degree is equal to J by 5 it will be equal means then we will equate those two so it will be solved. Or so I think as I am new to this topic.

[subtle imp]

Speak in pre-grad please

[subtle imp]

yep, that works

[sleek nimbus]

After you have changed the subject of the formula enter the numbers and get the results

[subtle imp]

lol

I was overcomplicating it

ty for your help

.close

[marsh citrus]

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[frail lodge]

for b), why can't I just directly differentiate?

[frail lodge]

Why must I use the limit def

So f_x = 0 = f_y = 0

[devout mauve]

because thats how the derivative is defined

[frail lodge]

but why can't I just differentiate informally

[devout mauve]

if f wasnt piecewise then you can skip this step by using the standard rules for differentiation

[frail lodge]

why is it different for piecewise

basically what my question is

[devout mauve]

those standard rules assume that for f(x+h) and f(x) the "formulas" are the same

but if they arent (as in the case of a piecewise definition) then your standard rules dont work

[frail lodge]

but in the question above, I showed that it was continuous everywhere

are you saying that single "hole" at (0,0) could potentially be non differentiable

[devout mauve]

f(0) could be defined as basically anything here. that surely has to matter for the derivative

[frail lodge]

so we can't apply differentiation rules because that one point

[devout mauve]

well you will see in (c)

[frail lodge]

so differentiation rules can only be applied if f is differentiable EVERYWHERE in iits domain?

[devout mauve]

differentiation rules can be applied for each piece of the piecewise definition

but not if your point is exactly at the boundary of one of those pieces

[frail lodge]

ok it says it is differentiable there

so suppose we verified this beforehand

then could we use differentiation rules?

[frail lodge]

but 0 IS the second piece/branch of the piecewise isn't it

or is it special for single-point branches

[devout mauve]

.

[frail lodge]

ok

[devout mauve]

this shouldnt be news to you. the same thing happens in one dimension

[frail lodge]

but if we verified the point is differentaible

then could differentiation rules be applied?

[devout mauve]

yes

[frail lodge]

Thanks!

[devout mauve]

well and then compute the limit for the derivatives

actually, wait

no

that doesnt work

[frail lodge]

wat

[devout mauve]

eg for x^2 sin(1/x) it is differentiable at x=0 but if you take the derivative by differentiation rules and then take the limit x->0 it doesnt exist

f(0)=0 piecewise def

[marsh citrus]

@frail lodge Has your question been resolved?

[marsh citrus]

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[small otter]

Can someone help me with these

[small otter]

Find the inverse of f(x)=x²-6x+8 and sketch its graph

<@&286206848099549185>

Is it possible that this equation have graph?

[twilit sable]

Sure thats a quadratic polynomial so the invers would have different branches
x=y²-6y+8
x-8=y²-6y
x-8+9=y²-6y+9
x+1=(y-3)²
±Sqrt(x+1)=y-3
y=±Sqrt(x+1)+3

It is convension to ussualy take the pricipled/positive branch of this function so i will take y=Sqrt(x+1)+3

[small otter]

Hmm

[twilit sable]

For the graph

U can graph it bcs it is a sqrt function

but there no singular invers to the stated function

it is divided into branches

[small otter]

$$x=y²-6y+8$$
$$x-8=y²-6y$$
$$x-8+9=y²-6y+9$$
$$x+1=(y-3)²$$
$$±Sqrt(x+1)=y-3$$
$$y=±Sqrt(x+1)+3$$

[elfin berry]

Jirou

[small otter]

How did you get the graph?

And where did you get the 9?

Nvm i found where to get the 9

[twilit sable]

I used a graphing calc but u can easily deterimine it using translations on the sqrt(x) function

Think of it as the sqrt(x) function shifted to the left by 1 bcs of the x+1 and shifted 3 units up bcs of the +3

[small otter]

I have a question when to use the sqrt?

[twilit sable]

You use the sqrt when u need to get rid of a x² term

sqrt is just a typing way of writing square root

[small otter]

So it's like cancelation the sqrt and ²?

[twilit sable]

yes

[small otter]

Can you teach me how to graph it?

[twilit sable]

sure

So for the sqrt(x) its an elementary function so u can just plug in values and get a general feel about how it looks

once u have the sqrt(x) graphed

u can start moving the graph

sqrt(x+a) means u shift the sqrt function a units to the left

and sqrt(x)+b means u shift sqrt(x) b units up

[small otter]

If it's sqrt the sign is like this ≥ instead of this=?

[twilit sable]

its the same idea its just you need to higlight the part of the graph thats effected by the sign

[small otter]

Oh but why use > instead of =?

[twilit sable]

its to signify that's its an inequality

but i dont think we use > in this question

[small otter]

Oh I forgot it's inequality

So the domain is like this?

x≥-1?

[twilit sable]

yes

[small otter]

What's next?

[twilit sable]

thats pretty much it u shift the graph according to the values in the answer

and then youl get the graph i gave you

[small otter]

There's no range?

[twilit sable]

well if you need the range its y>=3

[small otter]

How to get it?

[twilit sable]

you can argue that since sqrt(x) only produce positive numbers so the range must be 3+ a real positive number so it must be y>=3

[small otter]

Where did you get the 3?

[twilit sable]

its the minimum value of the function you can get it by plugging in -1 to the function to make the sqrt part = 0

[small otter]

Oh ok

Is there any x and y intercept in this?

[twilit sable]

there is a y intercep but thats easily found by just substituting x=0

there no x intecept

clearly by the range not even being able to hit 0

[small otter]

Oh ok thank you

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[marsh citrus]

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[marsh citrus]

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[lyric bay]

Hello

[lyric bay]

Can someone explain me the steps for this problem?

Like why they did these steps

[lone heart]

Do you know what RREF is?

[lyric bay]

Making the leading pivots 1

[lyric bay]

In a staircase form

No pivots on same level

[stoic saddle]

these ones specifically?

[lyric bay]

Yes

Ohh nice name change

[stoic saddle]

probably one of their motivations was to keep things in the integers

so they didn't just want to do R_2 - (3/2)R_1

[lyric bay]

So there are always multiple ways to solve rrefs?

[stoic saddle]

ys

yes*

the RREF can be arrived at through many many different routes

[lyric bay]

Okay

[lyric bay]

Do you know how the upper rref can be derived like the following?

Also, can I just multiply 1/2 for R_2 in the third step?

brb

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[distant peak]

hm ?

[willow owl]

ok so first of all you need to know the domain of the function

[willow owl]

he probably wants to solve for x

[whole jetty]

x-2

oh wait i alr know how to do

sorry for bother

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[willow owl]

wait

do you know how to solve

or know how to get domain of definition

[whole jetty]

i alr know how to solve, i was doubting on how to solve it because i didn't thought that c=0

[willow owl]

hm ok

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[olive sky]

i2 is found to be -5A, and difficuty in finding i5 and i4

help

[main idol]

#old-network for CS or EE server

[olive sky]

thanks

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[violet prairie]

where did i mess up here?

7y - 2z = 27 is from eq 1 + eq 3 and 17y - 4z = 69 is eq 1 + (2)eq 2

nvm

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[summer trench]

,, \sin(A+B) \times \sin(A-B) = \sin^2{A}-\sin^2{B}

[summer trench]

this is true, right?

[elfin berry]

!Yajat!

[main idol]

,tex .sum diff trig

[elfin berry]

riemann

[summer trench]

okay so we basically derived it from these