help-42

[hollow mural]

find common denomiator and then solve

[rustic blaze]

heres #14 again wtf is wrong

ignore the 4x

i eraded alot

erased

[hollow mural]

then if you don’t multiply top and bottom, multiply both sides of the equation with the produce of all 3 denominators

[rustic blaze]

cross multiplying

ans canceling out

[hollow mural]

how did the other side become 8

[hollow mural]

yea

[hollow mural]

nevermind

[rustic blaze]

whats wrong with it 😭

the answer is 26 because i looked it up

idk how they got it

[hollow mural]

yeah thsts seems correct

lm double check

[rustic blaze]

the answer is 26 😭

idk how to get 26 out of that

[hollow mural]

yea i got that

should be 2x + 2

highlighted part

[rustic blaze]

does the first 2 cancel out

[hollow mural]

yes

[rustic blaze]

how do those even multiply

[hollow mural]

?

[rustic blaze]

how do those multiple

multiply

you will atill have a demoninator

[hollow mural]

you get 5x + 40 instead of 10x + 80

[rustic blaze]

whwre does the x+2 go

whwre is it

it stays

[hollow mural]

yes but they all have the same denominator, so they cancel out

[rustic blaze]

it cant stay

x+2 has to be removed

[hollow mural]

what x + 2?

[rustic blaze]

and it doesnt get removed

[hollow mural]

your working is correct except for this highlighted part

[rustic blaze]

the 2 factors out which leaves 5/x+2

what do you do with x-2

it cant be there

[hollow mural]

idk what you’re saying

[hollow mural]

redo your wiring but change x + 1 to 2x + 2

you get 5(x + 8)

no x + 1

[rustic blaze]

@hollow mural

[hollow mural]

[hollow mural]

its -(6x + 6) not -6x + 6

[rustic blaze]

wtf is the problem

[rustic blaze]

huh

how do you get 26

please help me

[hollow mural]

^

[rustic blaze]

where did you get the -

[hollow mural]

where did you get it?

[wraith shuttle]

I think you're trying to solve this too fast without actually understanding what you're doing

Which is just gonna make your situation worse, and take longer

[rustic blaze]

so this

[wraith shuttle]

Js

[rustic blaze]

1:10 am

not slewping unti lits solved

until its

[wraith shuttle]

Yeah, but take your time

Rushing doesnt help

[hollow mural]

where did you get- from?

[rustic blaze]

it was a mistake

look at it this time

[hollow mural]

well there’s a negative

[hollow mural]

still wrong because you’re subtracting the other fraction

[hollow mural]

thsts the same thing you had before

[rustic blaze]

yes

whats the wrror

wrror

error

[hollow mural]

because you’re subtracting 3(2x + 2)

so it wouldn’t be what you have

this is being subjected

subtracted *

because it’s negative

in the equation

[rustic blaze]

@hollow mural

fianlly

i can rest

[hollow mural]

nice

[calm coral]

@rustic blaze Has your question been resolved?

[calm coral]

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[vale wraith]

could someone walk me through this/how do i proceed?

[formal tulip]

c would determine the value of the leading coefficient

So what would that be?

[vale wraith]

uhhhh

1?

[formal tulip]

perfect

[vale wraith]

how would i find e from there

[formal tulip]

Multiply the constant terms in every factor

Or find p(0)

Same thing in essence

[vale wraith]

im not following

🥲🥲

[formal tulip]

If you have p(x) = x^4 + bx^3 + cx^2 + dx + e, p(0) = 0^4 + b0^3 + c0^2 + d0 + e = e

So just find p(0)

From the factored form

[vale wraith]

?

[formal tulip]

Yes

[vale wraith]

ohhhhhh i see

thank you!!

i have a question regarding part b

(this is the answer key)

so i understand up until the 0=x^3-16x^2 + 79x -190

i know that the next step is synthetic division

but why do they use 10 as the divisor?

how do they know it’s 10

[bronze adder]

Its mostly guess and check when they used the rational root theorem

[vale wraith]

so rational root theorem is the p/q thing right

factor of last term/factor of first term

[vale wraith]

so why is the q +-190? i thought 190 was the last term

so it would be the p not the q

but for q it has +-190 beside it even though 190 is the last term

[calm coral]

@vale wraith Has your question been resolved?

[calm coral]

@vale wraith Has your question been resolved?

[calm coral]

@vale wraith Has your question been resolved?

[bronze adder]

Yes

[bronze adder]

You need to consider +- as while using it

If you think about why the rational root theorem is true, it should pop out why

[calm coral]

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[covert mason]

<@&286206848099549185>

[calm coral]

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[golden crater]

hi guys

I have a problem

[golden crater]

guys I am not understanding this
Degree 4. Roots of multiplicity 2 at x = -1/2
and roots
of multiplicity 1 at x = 6 and x = −2. y-intercept
at (0,18).
I did this question like 18=c(x-1/2)^2 (x-6) (x+2)
and solved for C

but apparently its wronf

its not x-1/2 its 2x-1 can someone explain why

<@&286206848099549185>

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[burnt cobalt]

for r'(t) i got <-psin(t), pcos(t), q>
for r''(t) i got <-pcos(t), -psin(t), 0>

since the formula for an is sqrt(|a|^2 - (at)^2), and at is 0, isn't the answer just p?

[calm coral]

@burnt cobalt Has your question been resolved?

[burnt cobalt]

<@&286206848099549185>

<@&286206848099549185>

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[burnt cobalt]

.reopen

[calm coral]

✅

[burnt cobalt]

<@&286206848099549185>

.close

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[amber bolt]

you need to know the secret number

[sharp summit]

what secret number?

heat capacity?

[amber bolt]

vaporization energy

[sharp summit]

o

so 644 x 2257

[amber bolt]

0.644

[sharp summit]

why 0.644?

.close

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[formal canyon]

I have 0 clue how to solve this problem I would really like some assistance

[leaden thunder]

cropping is your friend

[formal canyon]

I Took it straight on discord but thanks

[stray tundra]

that is missing information, i believe. How many hours per week do they work?

or is that what you're asked?

[formal canyon]

I’m doing algebra so I have to write a “let x =…”

[stray tundra]

okay, then they want you to determine payroll as a function of said hours worked weekly

said payroll is the sum of the payrolls of each person

[formal canyon]

So the variables are the hours worked

I’m going to just ask my teacher for clarification

.close

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[thorn parcel]

I understand how to get to the exponential function, but how do I do the linear function?

[golden panther]

instead of considering the general form of an exponential function y = ab^x, you should consider the general form of a linear function y = ax + b

from there the steps are very similar

[thorn parcel]

That makes sense but in the second problem the linear form is y = 21 + 21(x - 2)

Which is the answer the teacher gave us

And the linear form is in the explicit arithmetic formula

?

<@&286206848099549185>

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[fringe pumice]

I think I made a mistake in the circumcenter I just don’t know how and where

[fringe pumice]

<@&286206848099549185>

[calm coral]

@fringe pumice Has your question been resolved?

[hearty sage]

hello

[fringe pumice]

Hi

<@&286206848099549185>

[fringe pumice]

<@&286206848099549185>

[hearty sage]

do you need help

be quick im at school

[fringe pumice]

Yes I need help with checking the my work, I don’t think the circumcenter is correct

<@&286206848099549185>

[hearty sage]

ok. Tell me about the problem

[fringe pumice]

This is the problem

@hearty sage

[fringe pumice]

<@&286206848099549185>

[calm coral]

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[neat gorge]

This doesn't seem to be a satisfying explanation.

I've seen other explanations that are definitely reasonable.

For example, if the distributive property is to hold, we have:

-a(-b + b) = 0
which is: -a(-b) + -ab = 0

Which shows that -a(-b) must be a positive for the equality to hold.

Lang's explanation using his example:

(-1)(-1) = -(1(-1)) = -(-1) = 1

This doesn't seem to show anything to me. The steps are all obvious and sensible, but I don't see how it reveals anything about multiplying two negatives. It seems that we're saying:

"see, two negatives equal a positive, let me prove it: (-1)(-1) = 1. See, this shows that two negatives equal a positive."

I think I'm missing something.

He tells us to prove the general case with -(ab), but I run into the same problem.

To be clear, I know that two negatives equal a positive, and I've seen other more satisfying proofs of it.

I'd just like to understand what I'm not seeing from his explanation.

[calm coral]

@neat gorge Has your question been resolved?

[spare beacon]

It's hard to comment on it if you don't show what properties you are working with

[neat gorge]

I'm working through Lang's Basic Mathematics.

Up to this point, we only have the properties of addition and distribution.

[spare beacon]

which of those steps do you take issue with

[neat gorge]

None of them my only issue is:

(-1)(-1) = -(1(-1)) = -(-1) = 1

I do not see how it accomplishes what (I think) he says he wants to accomplish, which is showing that two negatives equal a positive. I do not see how the steps he provides in the picture reveal anything about multiplying two negatives.

All of the properties make sense. I'm only failing to understand the proof using his method.

[calm coral]

@neat gorge Has your question been resolved?

[calm coral]

@neat gorge Has your question been resolved?

[calm coral]

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[velvet osprey]

funny how they give you something called l but then ask about I, which they tell you nothing about.

[remote mural]

I think I is the integrated l

[calm coral]

@covert mason Has your question been resolved?

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@covert mason Has your question been resolved?

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@covert mason Has your question been resolved?

[still marlin]

damn u still on this problem

Maybe the square root of 2 is a bambooxle

Just there to make you think Taylor series won't work

idk if you ever bothered trying it out so I'll try it soon

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[grim imp]

How to go about solving this?
On 1 and -1, all of them make sense but sec and cosec don't have any values.

[steady hatch]

Isn't

, w arcsin(x) +arccos(x)

[potent lotus]

Results provided by WolframAlpha

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

Requested by ericsson6890

[steady hatch]

Ehh

[steady hatch]

Isn't f(x) = (3pi)/2?

[grim imp]

I'm talking bout sec^(-1) and cosec^(-1) —
Isn't there domain R - [-1, 1]?
And sin^(-1), cos^(-1) have domain [-1, 1]
It would mean sec and cosec doesn't have a value for any x that sin, cos has values for.

[calm coral]

@grim imp Has your question been resolved?

[still marlin]

wot

sec = 1/cos

If cos = 1

Then so does sec

So same angle should work

And inverse sec returns that angle value (in a specified range)

Same for arccsc

[grim imp]

ohk, seems I was mistaken bout the rang, domain of sec and cosec.

so that means, this question can be framed as sin^(-1) + cos^(-1) + tan^(-1) + cot^(-1) + 1/(sin^(-1)) + 1/(cos^(-1)) ?

[warm kiln]

hey friends , i cant seem to solve a problem can u help me out ? it says : proof that v is a solution for the diffrential equation (y'-2y=0) only if v+u is a solution for (y'-2y=xe^x) where u(x) = (ax+b)e^x

[calm coral]

@grim imp Has your question been resolved?

[grim imp]

Sound tad complex for me. Bet someone else can help u.
But this channel is under use right now. I'd recommend asking in another one.
#❓how-to-get-help

[warm kiln]

alr bet thx

[calm coral]

@grim imp Has your question been resolved?

[still marlin]

yeah

[grim imp]

ty for help.

.close

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[calm coral]

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[grim whale]

how is f(x) 1/x a continuous function

[grim whale]

😭

isnt 1/x discontinuous at x=0

o wait but the domain of 1/x dosent include f(0) so its still continuous....?

but i need to take my pen off paper

[remote mural]

the notion of continuity at a point $x_0$ only applies to the points of the domain for the function

[grim whale]

ohhh ok thanks :)))

[remote mural]

The notion of discontinuity does not apply only to points of the domain, though

you need to understand this distinction

[grim whale]

could u elaborate a bit on this...

[remote mural]

so a continuous function can have discontinuities, but not in points of the domain

[stoic oyster]

now it is true that saying 1/x is continuous on its domain doesn't say much
most interesting theorems about continuity require continuity on an interval

[grim whale]

but bc its not in the points of the domain the function itself is still continuous right

[grim whale]

we didnt get that far yet in class man 😭

[stoic oyster]

for example the intermediate value theorem is completely screwed up with 1/x around x=0

well you'll get there ^^

[grim whale]

that sounds hard i hope i dont ;-;

[grim whale]

could someone confirm this statement

:D

[stoic oyster]

i think alex is talking about the different types of discontinuities that exist

see https://www.youtube.com/watch?v=joewRl1CTL8&pp=ugMICgJmchABGAHKBR5vcmcgY2hlbSB0dXRvciBkaXNjb250aW51aXRpZXM%3D

you need the function not to be defined at the point to have an infinite discontinuity for example

oops I replied to alex

[grim whale]

ye but even tho there is an infinate discontinuity at x = 0 the funciton 1/x is still a continuous function is what he explained to me yes?

bc the point x = 0 is not in the domain of the function?

[grim whale]

i will be watching this thanks :)

[remote mural]

@grim whale

here is a small little exercise for you

[grim whale]

yes?

[potent lotus]

For the functions \[
\map f x = \env{dcases*}{
\hp - 1 &if $x \ge 0$ \\
 - 1 &if $x < 0$
}
\tss{and} 
 \map f x = \env{dcases*}{
\hp - 1 &if $x > 0$ \\
 - 1 &if $x < 0$
}
\]

[remote mural]

which one of them is continuous

[grim whale]

neither?

[remote mural]

are you sure?

[grim whale]

no ;-;

bc for the 1st one its still defined at x = 0

but for the 2nd one it isnt

wait whats the difference between a point being defined and a point existing

[remote mural]

what do you mean by that

[grim whale]

idk i hear those 2 termes being used and idk the difference

[remote mural]

Dang, even I didn't know that

@grim whale In that case I might've mislead you a bit in the previous help channel

[grim whale]

https://tenor.com/view/i-trusted-you-silco-arcane-you-betrayed-me-trust-gif-23705027

jk

loll

[remote mural]

I'm evil

[grim whale]

alex gone

he lef me

HE BACk

[remote mural]

the first one is not continuous because the limit in 0 does not exist

Alex you're a hero

[grim whale]

okok

i would worship alex

[remote mural]

the second is continuous because it is continuous in each point of its domain

💀

[grim whale]

OKAYYY

WAIT THAT MAKES SENSSEEE

[hot temple]

in the second, x=0 isn't in the domain so it doesn't count for whether it's continuous 😌

[grim whale]

THE DOMAINNN🙌

[hot temple]

everyone hates continuity

[remote mural]

I love continuity

[grim whale]

ik this is basically the same thing but can some math genius pls confirm above statement so i dont mislead when i offer the little knowledge i have to otehr people

😭

[remote mural]

Yes, if all points in the domain of the function are continuous then the function is continuous

[grim whale]

thank you

https://tenor.com/view/happy-feet-penguin-im-free-happydance-mumble-gif-11367015

ok thank ygs so much for help fr man ygs are lifesavers :))

bye :DDD

[remote mural]

thats why you shouldnt be confused, for when for example, someone says ln(x) is a continuous function

even if its not defined at x = 0 or something

[grim whale]

yasssss

[remote mural]

@remote mural Is sqrt a continuous function? f(0) is defined but the left limit as x approaches 0 of f(x) doesn't exist

[remote mural]

bye

[grim whale]

.close

[calm coral]

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[remote mural]

i have read up about this a while back

[calm coral]

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[grim whale]

isnt it continuous bc its continuous everywhere in its domain

[grim whale]

o shoot

[remote mural]

You can only talk about continuity on the domain where the function exists. The function doesn't exist for $x < 0$ so we cant talk about limit values for $x < 0$

[remote mural]

i believe

[grim whale]

W

im a math genious now

.close

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[weak rapids]

alex your profile is so neat

[remote mural]

thank youu

[calm coral]

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[azure willow]

How would I inverse laplace 2/(s^2+4)^2? I tried looking at the tables but i couldnt find anything in that form

[remote mural]

break it up into two

then use convolution theorem?

[azure willow]

how?

[remote mural]

[
\map F s \cdot \map G s = \f2{(s^2+4)^2}
]
for $\ds \map F s = \f2{s^2+4} \tss{and} \map G s = \f1{s^2+4}$

[remote mural]

i mean you can do it using convolution theorem surely, im just afraid i am overcomplicating this perhaps

been a while since i did Laplace

[azure willow]

Sorry if this question sounds too stupid but

[azure willow]

In here, does L(F(s)G(s)) = L(F(s))L(G(s))

i just got started with laplace transform so im sorry if this sounds stupid

[remote mural]

no it doesnt

[azure willow]

oh

does this formula help?

[remote mural]

if you have $\cmap{\LLL}{\map f t * \map g t} = \map F s \map G s$, then: [
\cmap{\LLL^{-1}}{\map F s \map g s} = \map f t * \map g t =\int_0^t \map f \tau \map g{t-\tau}\dd \tau
]

[azure willow]

does the * mean multiply?

[remote mural]

i dont think it applies here? i can only think of convolution theorem right now, sorry

[remote mural]

no

it denotes the convolution

[azure willow]

f(t) g(t) are unknown?

[remote mural]

thats what you are trying to find yeah

and then compute the integral

[azure willow]

did i set it up correctly?

sorry there should be a 2 infront of the L

[remote mural]

yeah

what is the inverse laplace of F(s) and G(s)

individually

[azure willow]

i dont know sorry 😭

[remote mural]

okay no worries

[remote mural]

you can use this then

im dumb

it does apply here

you have a = 2

but your numerator is 2 only

so you need to add a 2^3 term to the numerator

i.e. [
\f{2}{(s^2 + a^2)^2} \tss{is what you have right now, and you want to transform it to} \f{2a^3}{(s^2+a^2)^2}
]

[azure willow]

yeah

[remote mural]

what do you multiply by the numerator to get the 2a^3 at top

[azure willow]

a^3?

[remote mural]

yeah

but you need to preserve the equality

[azure willow]

a^3/a^3

[remote mural]

yes!

also

laplace transforms are basically integrals

with integrals, you can pull out constant terms

so you can pull out the 1/a^3

do you know what to do now

[azure willow]

yep

(sin2t - 2tcos2t)/8?

[remote mural]

yeah!

good job

[azure willow]

omg tysm

🙏

thanks for helping

.close

[calm coral]

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[red sapphire]

So I have to calculate exact value of arcsin(sin(-4)) and my question are:

• what is sin of integer (why not π or degrees, how to calculate that integers)?
• how to solve when problem is arcsin(sin(x)) and not sin(arcsin(x))?

[ivory pilot]

So you just don't know that
arcsin(sin(x)) = x

[fringe reef]

this is incorrect

[red sapphire]

sin(arcsin(x)) is not x, right?

[ivory pilot]

Huh ?

[ivory pilot]

Yes but under the condition
1 >= x >= -1

[ivory pilot]

I understand your point now

[fringe reef]

arcsin(sin(88888 pi )) = 0 ≠ 88888 pi

[ivory pilot]

That's not what I expected

I thought you would talk about
Arcsin(sin(130⁰)) ≠ 130

[fringe reef]

same thing

[red sapphire]

For arcsin(sin(-4)) I know answer is π - 2, but I don't know how to calculate that

[ivory pilot]

It is just because
Arcsin will output the primary acute angle only
Because sin(130⁰) = sin(50⁰)
So it will output 50⁰
So yeah arcsin has some little problems

[ivory pilot]

So you are now talking about radian right?

[red sapphire]

Yeah, I have to provide radians

[ivory pilot]

So for some angle x
We know that
$\sin(\pi-x)=\sin(x)$

[potent lotus]

Sherif Player

[red sapphire]

Ok, so it's some rule

[ivory pilot]

It is like an identity

Wait a min to send you a picture of it

It is not here lol

But you can see if you used sin(a-b)

Sin(π-x) = Sin(π)Cos(x) - Sin(x)Cos(π)

If you done it correctly you will get

Sin(π-x) = Sin(x)

@red sapphire Understood?

[red sapphire]

One moment, i'll write it on paper to see

[ivory pilot]

It doesn't need paper to do it
Sin(π) = 0
Cos(π) = -1

[red sapphire]

Right, but can sin(4) be transformed to pi form or it has to be calculated as number in range of 0;1

Ah, ok I understand now

Thanks

[ivory pilot]

Please wait

I don't think we can do it easily but to do it
First
Sin(-4) = -Sin(4)

You might ask why we did that
You will understand

-Sin(4) = -Sin(4-π+π)

-Sin([4-π]+π)
Now we just use the identity
Of Sin(a+b)
We will get
-Sin([4-π]+π) = Sin(4-π)
You know that π = 3.14 approximately
So this number will be less than 1
Which means that it will be less than π/2 which indicates that the angle is an acute angle

So we can say that
Arcsin(Sin(-4)) = Arcsin(Sin(4-π)) = 4- π

[red sapphire]

Oh, I get it

Thank you again for your time ❤️

[ivory pilot]

No problem
When you face a problem like that you have the table of identities that you can mess around to get the answer
You don't have to do all of that stuff if it is on ArcCos or ArcTan because they don't have these confusing properties of ArcSin

It is Just
ArcCos(Cos(x)) = x
ArcTan(Tan(x)) = x

Any questions?

[red sapphire]

Not really, thanks

[ivory pilot]

.close

[calm coral]

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[jolly anchor]

hello why in this i dont have to ckeck denominator different 0?
only root >= 0

[flat laurel]

What exactly is the question for ?

find domain?

[jolly anchor]

yup

i dont really understand this

x²-9>= 0

and what is other condition

[flat laurel]

that’s the only you have worry

[jolly anchor]

why?

i have denominator as well no?

[flat laurel]

Try setting denominator =0

What do you get?

[jolly anchor]

do i take the whole thing and do it

[flat laurel]

yes

the whole thing shouldn’t be equal to zero

what do you get?

[jolly anchor]

something is wrong with this

[flat laurel]

What did you get?

[jolly anchor]

x>= +-3

[flat laurel]

I mean have yiu se the denominator =0

$1+ \sqrt{x^2-9}=0$

[potent lotus]

.doc

[jolly anchor]

yeah

it makes

x²-9 >= 0

[flat laurel]

Not exactly

[jolly anchor]

just have to check condition of the root

[flat laurel]

There is no such real number which will make the equation true

1+ -1= 0 right?

[jolly anchor]

yup

yeah

so -9 cant be solved without

i

i get it ty

.close

[calm coral]

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[pliant coral]

!status

[calm coral]

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

[remote mural]

Idk where to start

[pliant coral]

ok

If a point (a,b) is on the line $8x-10y=C$, what does that mean?

[potent lotus]

CST (please ping when replying)

[remote mural]

X coordinate is a and y coordinate is b

[pliant coral]

well in terms of the equation

[remote mural]

Idk i don't think i get what you are saying

[pliant coral]

Ok

If the x-coordinate of the point is a and the y-coordinate is b, and the point is on the line…

[remote mural]

Should i just pick a point and insert the numbers?

[pliant coral]

yes

that’s basically it

[remote mural]

(5;0) for example

[pliant coral]

Yep

Fortunately for you C is already isolated

[remote mural]

40?

[pliant coral]

so you literally just plug in

Yep

It’s nothing more than a one-liner

[remote mural]

I always overcomplicate things

Thank you so much for the help

.close

[calm coral]

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[oak wraith]

I have attempted to answer this question and here is my current progress:

1. Triangle BDH is isosceles. Using this information, the measure of angle BDH has been calculated as 54°.
2. Drawing a line segment from G to H results in triangle GHI, which is isosceles. Angle GHI is 45° and the angle at vertex G is 90°.

However, I got stuck. I feel like I need to use angle FCG but I don't know how. I tried connecting other line segments but nothing seems to help me get closer to finding the angle.

[calm coral]

@oak wraith Has your question been resolved?

[calm coral]

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[topaz jasper]

Can someone explain how the chain rule derives this?

[leaden thunder]

Do you know chain rule

[topaz jasper]

yeah i know what the chain rule is but I am not sure how it's applied haha it just seems like they just apply it out of nowhere

[calm coral]

@topaz jasper Has your question been resolved?

[calm coral]

@topaz jasper Has your question been resolved?

[calm coral]

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[candid reef]

How would I find what value of x makes this expression undefined? (How would I find the non permissable values)

[storm stirrup]

x + (2/(x+3)) = 0

[candid reef]

What I'm confused by is that this expression can also be written as 2/1 divided by x² +3x + 2/x +3. I would then set the 1, x+3 and x²+3x+2 to 0, but obviously 1 can't equal 0

wait

I think I got it

I remembered you look at the original expression, thank you

.close

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[olive brook]

how do i find the number that the square rooted number is multiplied by?

[supple needle]

Is there a certain way you've been asked to approach problems like these?

[small matrix]

Anyone good at math?

[supple needle]

I would hope that most of us are. If you need help, you should go to #❓how-to-get-help

As this channel is occupied

[olive brook]

no there hasnt i just started the unit yesterday

[supple needle]

What have you learned about square roots so far?

[olive brook]

how to square root 1 and 2 digit numbers

but I understood that easily

[supple needle]

289 is a perfect square, so there is a number that when multiplied by itself will give you 289

[olive brook]

ok

[supple needle]

There are a couple ways we could approach this---I'm assuming you can't use a calculator for it

[olive brook]

no

sadly

[supple needle]

Often the first idea that people will try with new concepts like this is to just guess and check

If you know that 9 squared is 81 then you know that the square root of 289 is bigger than 9

[olive brook]

ok

[supple needle]

Alternatively, you could also list out the factors of 289 and see if you find two that are the same

This problem is harder to do that in, because its only factors are 1, 289, and it's square root which you're trying to find

[olive brook]

yeah i figured that

[supple needle]

If all else fails, try guessing and checking different numbers

If you want a hint, 20 is too high (20 squared is 400)

[olive brook]

thanks

.close

[calm coral]

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[golden crater]

hi

[golden crater]

does anybody know chem

[torn gorge]

no

[golden crater]

wondering If I could ask here

bruh

it doesnt hurt to ask

[torn gorge]

sorry you're right

[pliant hedge]

try the chem server in #old-network

[calm coral]

@golden crater Has your question been resolved?

[calm coral]

@golden crater Has your question been resolved?

[calm coral]

@golden crater Has your question been resolved?

[leaden thunder]

This is a math server

#old-network for an appropriate science server

.close

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[latent zinc]

Hi

[latent zinc]

To show that (u1,u2) have the same space as (v1,v2) is it enough to find a linear combination that makes it so:

x * u1+y * u2 = v1+v2

[golden crater]

ok

sorry mr boss mann

[calm coral]

@latent zinc Has your question been resolved?

[trail yarrow]

no, it would not be

a stupid counterexample would be that e.g. v1 = (1,0) and v2 = (2,0) do not span the same space as u1 = (1,0) and u2 = (0,1)

where 3\ * u1 + 0 * u2 = v1 + v2

what you do need to check is that a) v1 and v2 are independent and that b) you have x * u1 + y*u2 = v1 and x * u1 + y*u2 = v2 separately

(where the x and y can be different across the two equations in b)

(a) needs to be checked because u1 and u2 are independent

[calm coral]

@latent zinc Has your question been resolved?

[latent zinc]

Aight

[calm coral]

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[frosty marsh]

Guys I need help

[frosty marsh]

I used the sine law for this but Im still confused I got the wrong answer according to an ai math tutor

i got 8.78cm..

rounded to 8.8

[pure kayak]

,w (15.3/sin(77pi/180))(sin(34*pi/180))

[potent lotus]

Results provided by WolframAlpha

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

Requested by aldrnari_

[pure kayak]

hm

[frosty marsh]

yeah I got that

is it right

[pure kayak]

think so, unless im imagining the triangle wrong

let me draw it

[frosty marsh]

Can I show u what the ai did and the sine calc

[pure kayak]

the left one is nonsense
it implies the side opposite the angle D is DE

the right one, youve wrote them in the wrong boxes

34, 15.3, 77

ai is way off

as usual

youre right

[calm coral]

@frosty marsh Has your question been resolved?

[frosty marsh]

one more thing

I did the same thing here but cosine law

i got 64 degrees

rounded

this

is it fine?

[calm coral]

@frosty marsh Has your question been resolved?

[calm coral]

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[quartz adder]

have i wrote the piecewise function corrwectly?

[remote mural]

where does the 1125 and 4875 come from

[quartz adder]

point slope form

i plugged in a point to find the y intercept of the graph

so i could write the equation

[remote mural]

so it is randomly picked?

[quartz adder]

no, i use the end point of the first part as the point for the second part to solve

because they share that point

b/c its continuous

[remote mural]

but like tell me how did u get 1125 and 4875

what is the calculation

[quartz adder]

sure

so for part 2

we're given the slope 75

bc its 75 per t shirt

so we have y = 75x + b

bc we cant assume theres no y intercept

so we plug in a point that we know is on the that line

the point it shares with the previous line

(75, 90(75))

which is (75, 6750)

so you plug that in

and get

6750 = 75(75) + b

b = 1125

so the equation for that line is 75x + 1125

is that valid is that correct

@remote mural

[remote mural]

its not

[quartz adder]

why not

[remote mural]

first, 75(75) + b is not defined

because slope = 75 applies to 75<x<=150

second, the y-intercept should not be concerned here

[quartz adder]

why shouldnt the y intercept be taken into concern, if i ignore the y intercept then the equation of the graph is wrong

[remote mural]

so think about it like this

you want to buy 76 tshirts for example

for the first 75 t-shirts, it's 90*75

right?

[quartz adder]

yes

[remote mural]

for the 76th t-shirt, why should it be 75 + 1125?

you just spent an extra 1125 dollars for nothing

[quartz adder]

okay then what would the second equation be

would it for real just be 75x?

[remote mural]

75x for the price

for the total cost, should be 75x + 90*75

[quartz adder]

what no if i buy 76 shirts it would just be 75(76)

where are you getting total cost from

[remote mural]

75(76) means you are ignoring the price of the first 75 t-shirts

think of it as a function of time

not t-shirts

like for the first 75 seconds the speed is 90mph or sth

[quartz adder]

wait but then why isnt my original answer correct

wait

hold on

so the equation for the second line is just 75x + 90(75) ?

that doest make any sense

but if the equation of the line is just 75x then that means if you plug in 80

youd be paying 75 each for 80 shirts

which is not right

youre paying 90 dollars each for 75 shirts

then paying 75 each for 5 more shirts

so i still think my original answer makes sense

[remote mural]

let me draw a graph to show you what i mean rq

[remote mural]

tis the graph of ur func

since the cost of the 76th t-shirt must always be higher than the cost of the 75th t-shirt

and I was wrong when I said the second equation would be 75x + 90(75)

it should be 75(x-75) + 90(75)

which should make sense

[quartz adder]

my boy

[quartz adder]

this

simplifies

[quartz adder]

to this

[remote mural]

oop

my bad 😦

so ig it's correct

srry for wasting ur time teehee 😭

[quartz adder]

lmao its okay

u made the problem make more sense tbh

.close

[calm coral]

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[ivory steeple]

for part a) i tried $|z+1|=\sqrt{(a+1)^2+b^2}=\sqrt{a^2+2a+1+b^2}$ but dont know what to do from there

[potent lotus]

jstn.

[leaden thunder]

Do you know Euler's identity

[ivory steeple]

not realy

[leaden thunder]

Errr

Sorry I meant

[ivory steeple]

is there any way to do it without that

[leaden thunder]

,tex .demoivre

[potent lotus]

riemann

[ivory steeple]

i do not

[calm coral]

@ivory steeple Has your question been resolved?

[leaden thunder]

From |z|=1, find an equation with a and b

Use 1^2 =1

[calm coral]

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[calm coral]

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[steep vine]

How to solve this?

[austere surge]

What is y_n?

[steep vine]

I don't know

Mayne nth term

[austere surge]

n_th term in what?

[steep vine]

I don't know

[calm coral]

@steep vine Has your question been resolved?

[torn gorge]

we can't go further

[calm coral]

@steep vine Has your question been resolved?

[calm coral]

@steep vine Has your question been resolved?

[calm coral]

@steep vine Has your question been resolved?

[steep vine]

.close4

.close

[calm coral]

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[warped warren]

where is this 1 coming from in the last term?

[warped warren]

i can see that (24sec^2(u)) is being factored out of the function. I just can’t figure out where that 1 is coming from

[pure kayak]

the first 24sec^2(12t^2)

factor out and it becomes 1

[warped warren]

why is that being added to tangent here? why didnt the qty 48t^2 go to the tangent term

in the 2nd qty

[pure kayak]

a+ab=a(1+b) idk what you mean

they factored it out of each term

[warped warren]

yeah, im just having a harder time then usual seeing the commutivity with all these trig functions

i think i see it

ill keep practicing

thanks

.close

[calm coral]

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[celest trail]

y=√(|x-1|+2 ) how the graph shifts relative to y=√(|x-1|)

[leaden thunder]

,tex .transformation rules

[potent lotus]

riemann

[celest trail]

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[burnt cobalt]

so because being hopeful mean's you're only 10% likely to donate and the control is 13% likely donate, does this mean that being hopeful does not mean anything statistically significant?

[calm coral]

@burnt cobalt Has your question been resolved?

[burnt cobalt]

<@&286206848099549185>

am i correct in what i said?

<@&286206848099549185>

[burnt cobalt]

<@&286206848099549185>

[forest niche]

no i think that this data being hopeful does not lead to a statistically significant difference in the likelihood of donating to a black organization when compared to the control condition

[burnt cobalt]

wait so what does it mean then? Because being hopeful means that you are 10% likely to donate, but the control is 13%. so

doen't being hopeful mean that you are less likely to donate?

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[leaden thunder]

!original

[calm coral]

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.

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[bronze karma]

I need help with this

I know that I can use the derivative to find the slope of a tangent line at any point

f'(x) = -2/x^2

I plugged in x=1 into f'(x) to find the slope of the tangent line at x=1

but i dont know how to move on from there

[sterile escarp]

y2 - y1 = m(x2-x1)

[bronze karma]

well, m is 2

[sterile escarp]

if they are just asking for the instantaneous slope then tangent is fine, but judging from what u said abt moving on

[bronze karma]

and x = 1

what would y2 be?

[sterile escarp]

y = m(x - x2) + y1

evaluate f(1)

that will be ur y1

[bronze karma]

ok hold on'

[sterile escarp]

and x2 will be x = 1

[bronze karma]

y-(-2) = -2(x-1)

y+2 = -2x + 2

y=-2x

so its -2x?

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[bronze karma]

I know the derivative of y is 2x-2

then i evaluated f'(1) which is 0

I got 1,0

then i did y-0=0(x-1)

and got y=0

as the tangent line

is this correct?

[pure kayak]

no

f'(1)=0

[warm warren]

You need to plug the 1 into f not f’

[pure kayak]

f(1)!=0

[warm warren]

your tangent line passes through the tangent point at x = 1 wrt to f

f’ can look nothing like f or the tangent line

f’ is just the slope

[remote mural]

@bronze karma do you still need help?

[calm coral]

@bronze karma Has your question been resolved?

[calm coral]

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[crimson swift]

Prove that EF is parallel to BC
A The starting point of the circles

[calm coral]

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[final swallow]

Does anyone know the exact name of this topic

I’ve been trying to find examples to practice but it’s to no avail and I’ve done all the ones in the book

[final swallow]

i googled that and it doesnt show anything like these ones

[weary wyvern]

its just algebra with an emphasis on proofs

[final swallow]

ok thank you

[weary wyvern]

if you want textbooks, there are a couple

[final swallow]

ok thank you very much

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[chilly aurora]

wait

so

how do I go about solving this

[static summit]

Since 2x-1 is a factor, you can use remainder theorem

[pure kayak]

this channel is going to disappear soon

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[clear delta]

!status

[calm coral]

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

[remote mural]

Solve the following inequalities algebraically. Confirm your answer
with a graph.

So I have gotten to x^2-2x-x but I wasn't able to find two factors to find it's roots

[clear delta]

x^2-2x-x

1. there are two sides to the inequality; where's the other side?
2. you're missing a power somewhere, this doesn't make sense

[remote mural]

and obviously that equation is problematic because if I subbed it into the -b+/- sqrt b2 -4ac/2a formula I'd get fractions

[remote mural]

I'll break down mysteps

first I moved all the factors from the left side to the right

then I found that x-0 is a factor

then I did synthetic division

and ended up with that

[clear delta]

several things

1. there are two sides to the inequality; where's the other side?
2. you can't just divide both sides by x, or x-0; you need to preserve it as a factor, so you'll end up with x(stuff)
3. you didn't quite do your synthetic division right

[remote mural]

1. yeah I'm aware but tbf it probably doesn't matter because right now I'm just doing algebra
2. I subbed in 0, because doing so would make the function = to 0 which means 0 would be a factor in this case
3. I think i messed up on the step before which is why I did synthetic division incorrectly

[clear delta]

you need to keep both sides, it's like keeping units

or else you'll get confused whether you have e.g. 0 < stuff or stuff < 0

and yes i think you made an error combining terms

[remote mural]

okay, then in that case my thing would be $x^3-2x^2-x \leq 0$

[clear delta]

no it wouldn't

[remote mural]

I agree, I'll try that again

[clear delta]

\leq, by the way, if you want it (also \geq)

[potent lotus]

RecRio

[remote mural]

okay

my answer is different this time

probably the right one

$x^3-2x^2-3x\leq0$

[potent lotus]

RecRio

[remote mural]

x-0 is still a factor in this case

[clear delta]

yep

[remote mural]

since subbing in 0 would mean the eq. equals 0

[clear delta]

(and since everything has a factor of x)

[remote mural]

now we're working

[clear delta]

$\polyhornerscheme[0]{x^3-2x^2-3x}$

[potent lotus]

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

[remote mural]

$x(x+1)(x-3)$

[potent lotus]

RecRio

[clear delta]

yep

weell

$x(x+1)(x-3) \leq 0$

[potent lotus]

Haylsune Miku

[remote mural]

yup, would 0 be included in the numberline in this case?

[clear delta]

yes of course; x=0 will make the expression on the LHS equal to 0

[remote mural]

okay, that makes sense then

$xe(10,inf)$

[potent lotus]

RecRio

[clear delta]

what?

oh

do you mean $x \in (10, \infty)$?

[potent lotus]

Haylsune Miku

[remote mural]

yeah

[clear delta]

oh then that's wrong

[remote mural]

hm yeah it shows wrong too

on the answer key

[clear delta]

when are each of those factors greater than / less than 0?

[remote mural]

when its x<-1, -1<x<0, 0<x<3, it's less

x>3, it's more

[clear delta]

make a chart

for each of the three factors

when are they greater than 0? when are they less than 0?

ie when are they positive / when are they negative

what does that mean for their product?

remember that negative times negative = positive

[remote mural]

my chart looks somewhat like this

on the left, vertically stacked

x
x+1
x-3
f(x)=

on the top, horizontally stacked

x<-1, -1<x<0, 0<x<3, x>3

[clear delta]

     <-1  | -1<0  | 0<3  | 3<
x     -   |   -   |  +   | +
x+1   -   |   +   |  +   | +
x-3   -   |   -   |  -   | +
f(x)

[remote mural]

Ohhhh

I see

I picked 1 as a term

for -1<x<0

lol

ok so

yeah our thing is the same then

[clear delta]

so what should the f(x) line read?

[remote mural]

-,+,-,+

[clear delta]

👍

so now we can answer the question of when does f(x) ≤ 0

[remote mural]

in that case, xe(-inf,-1)U(-1,0)

[clear delta]

uh

why 10

[remote mural]

that was a term I picked

lol

[clear delta]

     <-1  | -1<0  | 0<3  | 3<
==========+=======+======+===
x     -   |   -   |  +   | +
x+1   -   |   +   |  +   | +
x-3   -   |   -   |  -   | +
f(x)  -   |   +   |  -   | +

so when is f(x) negative or 0?

[remote mural]

-1<0 or 3<

[clear delta]

those are + signs

[remote mural]

wait a minute

we're looking for

less than or equal

oh ok

less than -1 and 0<3

[clear delta]

ok, and what about at those points?

like what if x = -1

would your original inequality be true?

[remote mural]

it would, since it's less than or equal to 0