precalculus

[uncut mulch]

that gets you:
**-**4sqrt(6))/9 + 3

since the stationary point is positive/negative sqrt 2/3
x-coords of the stationary points

,w evaluate -4sqrt(6))/9 + 3

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by ramonov#0921

[delicate rivet]

where did the negative come from?

[uncut mulch]

the question is how youre getting a positive

[delicate rivet]

Heres my working, its really confusing but i basically found the stationary points and then tried to sub it ?

oh shoot

[uncut mulch]

bottom box

[delicate rivet]

nvm

thank you, that's my bad

[grim sand]

Can I get help?

Or is there a convo going on

[willow bear]

there seems to be no convo going on here at the moment

[grim sand]

Oh ok

What am I looking for 😔 I’m lost

@echo wagon can u help me

Is it (3,0)?

Can anyone help

[novel cargo]

there is an interval where the graph decreases

between 3 and 5 on the x axis the graph has a negative slope and the y value decreases as you move forward

is that what you had in mind?

[echo wagon]

What does decreasing mean @grim sand?

[grim sand]

Going down

[delicate rivet]

it should be the interval (3, 5)

as that's when the gradient is <0, and make sure the brackets are round since the points aren't included

[grim sand]

Huh

Is this a different convo or for me..?

[novel cargo]

for you

[echo wagon]

It's for you, lol

[grim sand]

Oh

[delicate rivet]

if you look at the cubic graph, when x is < 3 the graph is going up yea?

[grim sand]

@echo wagon plz translate 🥺

[echo wagon]

Since you say it has to go down, where is it going down?

[delicate rivet]

same with when x > 5, the only time the gradient/slope is negative is between 3 & 5

[grim sand]

It’s going down at 3 on Y axis

Yes

[echo wagon]

At 3 on y axis ? What does that mean?

[grim sand]

same with when x > 5, the only time the gradient/slope is negative is between 3 & 5
@amy.ames.aims#0827 I’m sorry but I don’t understand..

[echo wagon]

Can you colour in the part that is going down?

Like edit the picture

[grim sand]

Oh

Where is the shade

[echo wagon]

Yeah exactly

Now if you had to describe the x coordinates of that part of the graph, how would you do it?

[grim sand]

0

[echo wagon]

0?

[grim sand]

Cuz the line start at 3,0

[echo wagon]

There's literally no point on the entire graph where the x coordinate is

0

At least the visible part

[novel cargo]

a line starting at 3.0 has x coord 0?

[grim sand]

I’m confused

[echo wagon]

(3, 0) which is the x coordinate?

[grim sand]

The 0

[echo wagon]

No

[grim sand]

No

The 3

[echo wagon]

Yeah

[grim sand]

And the Y is 0

[echo wagon]

You really shouldn't get that wrong anymore

And at what point does it stop going down?

And what is the x coordinate of that point?

[grim sand]

At (5,-4)

5

[echo wagon]

Yeah, and x coordinate is?

Yes

[grim sand]

Why are we only looking for X

What about Y? It’s lonely

[echo wagon]

So do you agree it is going down for 3 < x < 5?

[grim sand]

Yes

[echo wagon]

That's a fair point. But when we consider when a function has a specific property, we look at which part of the domain it has that property

And the domain is the x values

So how would you write the interval of the x values with 3 <x < 5?

[grim sand]

(3,5)

[echo wagon]

So that's the answer

[grim sand]

Oh

Oki

[echo wagon]

That's the part of the domain where the function is decreasing

[grim sand]

So when we looking for when the function is decreasing

We only look for domain values?

What if it’s increasing

[echo wagon]

Well, everywhere else

You always look at the domain

If you ask WHERE DOES A FUNCTION HAVE SOME PROPERTY then you answer with a subset of the domain

One reason for this is that a function can have multiple points with the same y value, so if you refer to the y values, it isn't clear which point you are talking about

Whereas every point has a different x value

[grim sand]

Oh bet

Would I have to graph number 2 first?

(-2,4)

Is it right

[echo wagon]

You don't know about derivatives yet? If not, yeah, graph it

[grim sand]

What are derivatives

[echo wagon]

Lol, so yes, graph it

[grim sand]

I graphed it and looked at Domain only and got (-2,4)

[echo wagon]

Well, I'll trust you then

I didn't graph it so idk

[grim sand]

Whut

Ur not supposed to trust me

I’m the idiot here

[echo wagon]

Show me the graph then

[echo wagon]

Yeah, looks right

[grim sand]

How do I find limit

My teacher was supposed to teach me this but he doesn’t explain he just does the math

Im dum
@235711131719232931...#8067 it’s ok I’m dumb too

[delicate rivet]

My teacher was supposed to teach me this but he doesn’t explain he just does the math
@grim sand oof that's rough

[grim sand]

Yes so I’m lost

[viscid thistle]

My teacher was supposed to teach me this but he doesn’t explain he just does the math
@grim sand That's not good, at all

[grim sand]

I just copy the work he does

And not understand any of it

[delicate rivet]

do you have a text book you can read through or anything?

[grim sand]

No it’s online so it’s even worse

[delicate rivet]

i can give you my notes if you'd like, i think we're doing the same course

[grim sand]

Pre calc?

Senior year

[delicate rivet]

i have no idea im australian ://

[grim sand]

Oh

U understand these things tho?

[delicate rivet]

it's yr 11 maths, but your questions look similar to what i did last term

[grim sand]

I’m year 12 math

[delicate rivet]

yep p much,,, i'm little slower since im yr 10 tho ://

[novel cargo]

get a precalc book, I'm doing Larson's. Good IMO

[delicate rivet]

I’m year 12 math
@grim sand the questions look the exact same O-O

[grim sand]

Yes

I was supposed to take this in year 11

But

My counselor said no

So I had no math for 2 years

[grim sand]

So.. how do I find limits

[delicate rivet]

here's the first part,,, i'll put together the answer sheet

[grim sand]

Oh cool it tells me limits

[delicate rivet]

uh depends on the denominator

can i see the question?

[grim sand]

Ok

[novel cargo]

I have f(t) = 0.75 * cos((2 * pi * t)/3)

someone please check

[grim sand]

i wish i could help

[delicate rivet]

do you want the explanation or just the answer?

[grim sand]

but i dont even know what a wave motion is

explanation

[novel cargo]

don't worry Jupiter

[grim sand]

i like learning not a straight answer

[delicate rivet]

ok sure;

[novel cargo]

after two years of not doing math it will take a few painful weeks but then you'll feel much better

just don't give up and be consistent with your daily practice

[delicate rivet]

ok so with limits there are 3 different types of 'cases'

1. where x is a constant number, (for example f(x) = 2)
2. where the function has x terms but has a numerical denominator (eg 2x+3 --> the denominator is 1 in this case)
3. where the function has x terms but has 0 for a denominator

can you tell what 'case' yours is @grim sand ?

[grim sand]

wait i lost my website

ok i found it

its 2

[delicate rivet]

yep,

so the next step would be to expand

[grim sand]

[delicate rivet]

[grim sand]

how do you expand

[delicate rivet]

i mean factorise

sorry my bad ://

[grim sand]

how do u factorise

hop on vc and help? or just type

[delicate rivet]

you take out 'like/similar terms' and group them together

i'm sorry its nearly midnight here,,,, i don't want my mum to whoop my ass

just ask away tho, i have plenty of time

[grim sand]

3x^-x

what do i take away

the x?

[delicate rivet]

yes!

[grim sand]

so now its just 3^2

[delicate rivet]

uh, not quite

it'd be x(3x - 1)
because 3x^2 is equal to x * x * 3
and -x is equal to -1 * x

[grim sand]

what does factorise mean

it just means breaking the function apart?

[delicate rivet]

it means to take out any common factors between your terms

[grim sand]

@echo wagon plz dumb it down for me

guess luna is afk

so whats the answer to the equation? @delicate rivet

[delicate rivet]

skdjhk i'm sorry i keep disappearing,,

factorise and sub it the small number at the bottom

wait is it an infinity sign?

[grim sand]

this one doesnt have small number at the bottom

[delicate rivet]

ok bad wording, the little symbols underneath 'lim'

hmmm,,, i don't think i'm the best at explaining this for you,,,

[grim sand]

what does that -> even mean

im a visual learner

[stuck lark]

@delicate rivet i recommend staying in alpha so your attention isn’t divided while you’re getting help

[hallow thunder]

that means that x is approaching something

[delicate rivet]

@delicate rivet i recommend staying in alpha so your attention isn’t divided
@stuck lark yea,,, that's really on me i took up too much

[grim sand]

if ur getting help just ignore me

i dont wanna disturb with ur learning

[stuck lark]

i point it out mostly bc it’s rude to the helper when you’re not totally focused

[grim sand]

yes that is correct

So can someone free help me 🥺

How do I solve limits

[stuck lark]

not much computation needed, just note the polynomial degree & lead coefficient, then recall corresponding end behavior

[grim sand]

What

I’m dumb so I don’t understand any word u just said

[stuck lark]

google: polynomial degree, lead coefficient, end behavior (of polynomial functions)

and it’s really for jupiter to look this up (leave it jed)

[hallow thunder]

i was going to suggest how i solve them

nvm

[grim sand]

Ok

What do I do after searching it up

[stuck lark]

study it

[grim sand]

This isn’t helping

Cuz I can’t ask it questions

[stuck lark]

once you read and try your best to study, then you can come back and ask questions

[delicate rivet]

i point it out mostly bc it’s rude to the helper when you’re not totally focused
@stuck lark You're right, and I apologise. I bit off more than I could chew for a short period of time

[stuck lark]

thanks @delicate rivet but the apology should go to anubis

and jupiter if you feel like you can attempt the question after studying then here, to start you off

not much computation needed, just note the polynomial degree & lead coefficient, then recall its corresponding end behavior

[grim sand]

What is computation

Actually

[delicate rivet]

computation is referring to your working,

[grim sand]

Can someone tell me what this type of equations are called so I can watch YouTube videos

The limits and stuff

[delicate rivet]

like it's saying not a lot of thinking is required

have you tried khan academy?

[grim sand]

I hate that

It makes me more confused

[delicate rivet]

like the app?

[grim sand]

Tried it freshman year it didn’t work out well

[delicate rivet]

i'd suggest your teacher then, or a tutor since they can help you with specific questions you have

[grim sand]

What is this type of equations called

My teacher can’t teach so he’s useless

[sour wagon]

@grim sand take one word in what @stuck lark wrote at the time and try to unwrap it and see what it means. You kind of have to go through this because mathematicians use these to somehow make sense of it all. (Imagine some numbers being written in arabic, some in Kanji and some in roman numbers, we need common language for them)

[grim sand]

Oh

So it’s like different math languages but we need a specific one from each to combine to get to know the one I’m looking at?

[sour wagon]

well no but your example is even better. Yeah, each formula or even sentences like the one the guy wrote to you, they represent a bag of math that will solve your problem

but you first have to go through the trouble of understanding them

[grim sand]

Okay bet

[delicate rivet]

My teacher can’t teach so he’s useless
@grim sand surely theres more than one math's teacher in your school? what about the math's co-ordinator?

[grim sand]

Got any videos that are spot on or do I have to look for myself

[delicate rivet]

idk, i'm sorry mate

[grim sand]

@grim sand surely theres more than one math's teacher in your school? what about the math's co-ordinator?
@amy.ames.aims#0827 idk what that is but we only got 1 pre calc teacher here

And it’s online so it’s harder

[delicate rivet]

ooo, what about the other people in your class?

you could always shoot him an email asking for 15 minutes of one on one?

that's what worked for me when I fell behind a bit

[grim sand]

Talking to him is a no man I don’t understand anything he says

And 90% of my friends r as clueless as me

[delicate rivet]

what about the people doing well? would they be willing to help?

[sour wagon]

I think I used to be in similar condition when it comes down to education as you @grim sand where are you from?

[grim sand]

NC

North Carolina

Bum teachers

[delicate rivet]

that's in america right?

[grim sand]

Yes

[sour wagon]

oh I am from Eastern Europe

well the world is a small place 😄

[delicate rivet]

ohh k tyy

[grim sand]

what about the people doing well? would they be willing to help?
@amy.ames.aims#0827 no clue I’m too scared to ask them cuz idk them

[delicate rivet]

hmmm first place i'd start is factorisation videos then?

[sour wagon]

come on man that's at least 500% better and faster to get something if you can talk to someone who knows it and can talk to you in person

[delicate rivet]

you need to have a strong grasp of algebra before you can tackle any of the graphing stuff

[grim sand]

I don’t know how to get in contact with them

[delicate rivet]

^^ what allione said

[grim sand]

We use zoom and zoom doesn’t let me email them

[delicate rivet]

email, zoom, stay behind after class

dm them,

[grim sand]

U can dm from zoom?

[sour wagon]

go to them when you see them goddamnit they'll might even thank you for asking for help

[grim sand]

go to them when you see them goddamnit they'll might even thank you for asking for help
@sour wagon 😔🤚🏻

[delicate rivet]

U can dm from zoom?
@grim sand yes? i think so

[grim sand]

Idk it’s my first year using zoom

[delicate rivet]

click alt h / open chat and toggle down for the person you want

idk either man i only use it for co-curriculars,,, we use microsoft teams

[grim sand]

My pre calc period ended 2 hours ago I’ll do it tmr hehe

[delicate rivet]

2hrs? skjhkj aight u got the rest of the day to start learning the basics to differenciation calc

Glll

[grim sand]

Thank u

I’m learning polynomials rn

[delicate rivet]

like on your own or in class?

[grim sand]

On my own

[delicate rivet]

oo thats a good start,, just keep at it

[novel cargo]

trying to plot this in desmos

what am I doing wrong?

[willow bear]

use another letter for the input of f and z

and then in the table put f(g) and z(g) for your columns

[novel cargo]

wow

thanks

it shows that tan(x) is just pi/2 late compared to -cot(x)

[austere void]

are you sure it isn't pi/2 early? 😃

[jade iris]

Im just having a brain fart bc this is old material to me but how would you find the y=ab^x form of the exponential function f(x)=5^x8^x ( I know the answer a=1 and b=40 I dont understand how to work it backwards)

[blissful ridge]

$l^k\cdot m^k=(l\cdot m)^k$

[obsidian monolith]

The Godfather:

[blissful ridge]

@jade iris

[jade iris]

@blissful ridge why is a=1?

[blissful ridge]

Use the identity, I gave you and tell me what you get?

[jade iris]

40^k

[blissful ridge]

$40^k=1\cdot 40^k$

[obsidian monolith]

The Godfather:

[jade iris]

is it because 1 is considered an initial value so there isnt but the intial value cannot be zero which would the expression not valid?

@blissful ridge

[blissful ridge]

You can multiply 1 so any given expression and it won't change the value

[jade iris]

ah ok so what would be an example of a would not be 1 @blissful ridge

[blissful ridge]

$2\cdot 39^k$

$2\cdot 39^k$

[obsidian monolith]

The Godfather:

[jade iris]

so now a=2 and b=39

[blissful ridge]

Yeah

[jade iris]

Yep get it now @blissful ridge thank you!

[lucid owl]

Y=1/3f(1/2x+1)

I got it wrong what did I do?

[viscid thistle]

Parentheses?

[hushed sorrel]

i think it looks correct

[viscid thistle]

If you typed it in exactly like that then it’s not right

(1/2)(x+1)

[regal kernel]

A cubic function generally has the form
f(x) = ax^3 + bx^2 + cx + d.
If we know that for some x-value x = p we have
f(p) = 0,
then it must be true that x − p is a factor of f(x). Since we are told that
f(3) = 0,
we know that 3 is a factor

@viscid thistle Can you help verify the last statement of "3 is a factor"?

[viscid thistle]

Why was I tagged with this question

[regal kernel]

you might be able to help me since you have, "advanced" as a tag

[viscid thistle]

Can you... not tag people without a good reason

[regal kernel]

... alright man

[swift flume]

doesnt it mean f(3) = 0?

like, (x-3) is a factor of the function

[regal kernel]

yes it does mean that f(3) = 0

i added the bolded part to the statement

[swift flume]

yeah

[regal kernel]

do you think 3 is a factor?

or what are they trying to ask there

[swift flume]

I'm not very good at this either but I think that means

(x-3) is a factor of the function

like, if you divide the equation by (x-3) the remainder would be 0

cuz of the factor theorem

[regal kernel]

hmm alright

so instead of 3, it would be more accurate to say, "x-3 is a factor"?

[viscid thistle]

Yes, this looks like a restatement of factor theorem

But for a particular value of 3

[regal kernel]

thanks

[viscid thistle]

And yes x-3 is a factor sounds better to me

[willow stream]

Hey can someone help explain to me an identity

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

[obsidian monolith]

Dozenford:

[willow stream]

wouldn't it be instead

$1+cos(2x)/2$

[obsidian monolith]

Dozenford:

[willow stream]

Okay

This is the pdf I am using

for trig identities

[lime bolt]

well i guess that does also work

[willow stream]

which is easier though

[lime bolt]

they are equivalant anyway

[willow stream]

I figure sin^2

[lime bolt]

actually u need to divide the 1 by 2 as well

[willow stream]

Yes my bad

[lime bolt]

do u know the expansion of cos(2x)

[willow stream]

2cos(x)?

[lime bolt]

no.....

[willow stream]

oh the formula

1/2 + cos(2x)/2 ?

[lime bolt]

no that isnt the expansion

[willow stream]

Can you explain it to me please

[lime bolt]

it is cos^2 -sin^x

[willow stream]

Oh

[lime bolt]

do u know expansions of sin(x+y) and cos(x+y)

[willow stream]

$2sin((x+y)/2))cos((x+y)/2)$

[obsidian monolith]

Dozenford:

[willow stream]

To be honest, off the top of my head I do not

I can off of a formula sheet but that isn't exactly knowing it

[lime bolt]

why do u keep using these ones

this is a corollary of the main ones

[willow stream]

This is the file I was going off of

Is there a better resource?

if you don't want to open the file which I understand i'll send a screenshot of it

[lime bolt]

it is here

[willow stream]

ohh I see

I see my error with the org problem

I was supposed to use the Pythagorean identity

[lime bolt]

you should prove that identity

using the definitons if sinx and cosx

[willow stream]

Okay so for instance: $If sin^2+cos^2 = 1 then cos^2 = sin^2 - 1$

[obsidian monolith]

Dozenford:

[lime bolt]

do you know this picture

[willow stream]

Yeah the unit circle

[ember crane]

I have a question, is this function exist?

[weak sierra]

@ember crane well cot = 1/tan so yes

[ember crane]

but the square part is also applicable right?

[viscid thistle]

yeah

@ember crane

from cot(x)=1/(tan(x)) you can square both sides to get to that conclusion

@ember crane

$cot(x) = \frac{1}{tan(x)}$ then we have, $cot^{2}{x} = (\frac{1}{tan(x)})^{2}$ then we get $cot^{2}{x} = \frac{1}{tan(x)}\frac{1}{tan(x)}$ then finally, $cot^{2}{x} = \frac{1}{tan^{2}{x}}$

[obsidian monolith]

Sasuke Uchiha:

[viscid thistle]

...

[ember crane]

wow.... thnks

[tardy ridge]

why does 1 question need to be answered 3 times.

[viscid thistle]

Yeah like

I have no idea

https://tenor.com/view/wink-dog-animal-pet-gif-4800447

why does 1 question need to be answered 3 times.
2 times

[ember crane]

nah, its 3 XD

[viscid thistle]

https://tenor.com/view/ok-sorry-homer-simpsons-gif-9894844

Also

\tan(x) \sin(x) \cos(x)

[ember crane]

1 from CEO of your data, 1 from Al3dium aand one from Sasuke Uchiha

yes sir, thnk u very much 😄

[viscid thistle]

Ceo of your data didn't answer the "but the square part is also applicable" lol

So it's 2

but i explained with the most detail 😎 me acting liking that matters

...

[ember crane]

lol XD

[viscid thistle]

Someone ban him

now

wanna chess battle ✊

[mighty onyx]

$4\log x-3\log(x^2+1)+4log(x-1)$

[ornate wolf]

am I right to start off by doing:

2+4i = 2+ai / 4+bi ?

[obsidian monolith]

Albot1288:

[mighty onyx]

any one know how to rewrite this into a single logarithm?

[uncut mulch]

power law and then sum/difference to product/quotient

[mighty onyx]

huh?

[uncut mulch]

those reference the log laws you would need to apply

[mighty onyx]

so I don't just simplify it?

[uncut mulch]

simplificty is subjective here

[mighty onyx]

I think I should rewatch the lecture tbh

[uncut mulch]

$a\log(b) = \log(b^a)$ would be the power law

[obsidian monolith]

ramonov:

[uncut mulch]

which you should apply to all your terms if you intend to combine them into a single log

as the other law(s) you would need to apply are: \
$\log(a) + \log(b) =\log(ab) \
\log(a) - \log(b) =\log\br{\frac ab}$ \
(for $a,b > 0$)

[obsidian monolith]

ramonov:

[mighty onyx]

so for mine it would be $x^4+x^6+1^3+x^4-1^4$ then simplify it. Log is not part of the final answer he says.

[obsidian monolith]

Albot1288:

[uncut mulch]

no

that's not how exponents work

$\fdream$

[obsidian monolith]

ramonov:

[uncut mulch]

and applying the power law doesn't make the log dissapear

[mighty onyx]

I think he means after it's simplified into a simple logarithm don't put the log into the answer bar

like if it's $logaB+C$ just put $aB+C$

[obsidian monolith]

Albot1288:

[uncut mulch]

um put proper parentheses please

regardless you're applying the proper steps in combining the logs

[mighty onyx]

ok

[uncut mulch]

refer to the power law above and tell me what you have after applying it

don't bother expanding them btw. leave it in exponent form.

[low owl]

I'm sitting here trying to memorize the unit circle lol my brain is saying no

[viscid thistle]

Don’t try to memorize it

try to memorize the methods that created it

So you can logic your way through piecing it together yourself

[low owl]

will do

[mighty onyx]

I did it like this but how do I put that into typed form?

Nvm got it

$(x^4(x-1)^4)/((x^2+1)^3)$

[obsidian monolith]

Albot1288:

[delicate rivet]

I got (5x^3/2)/2, is that right? and how would you simplify it even more?

[blissful ridge]

Looks alright to me
And simplification is subjective

[uncut mulch]

[delicate rivet]

how would you simplify it? -ty btw

[uncut mulch]

\verb|(5x^3/2)/2| reads as $\frac{\br{\frac{5x^3}{2}}}{2}$

[obsidian monolith]

ramonov:

[uncut mulch]

but if you intended $\frac{5x^{\frac32}}{2}$ represented by \verb|5x^(3/2)/2| in plain text, that would be simple enough

[obsidian monolith]

ramonov:

[delicate rivet]

hmm yea i understand, but are there any that match one of the answers?

i think i'm simplifying it wrong or something,,,, i can't get their answer

[uncut mulch]

exponent - radical laws

what would x^(3/2) be as a radical?

[novel cargo]

x^1 * sqrt(x) = x^1 * x^1/2 = x^3/2

[delicate rivet]

sorry but what's a radical?

oh nvm i had to look it up

x^(3/2) = (root x)^3 yea?

x^1 * sqrt(x) = x^1 * x^1/2 = x^3/2
@novel cargo ohhhhh thank you!

[uncut mulch]

[novel cargo]

true

[delicate rivet]

how do we know when the gradient of a graph is defined?

[willow bear]

when the function being graphed is differentiable

i.e. when the limit $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ exists for whatever values of $x$ we care about

[obsidian monolith]

Ann:

[novel cargo]

what is the gradient of the graph?

[delicate rivet]

i think it's slope in america?

[novel cargo]

ahh, thanks

[delicate rivet]

i.e. when the limit $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ exists for whatever values of $x$ we care about
@willow bear ohhhh thank you

[obsidian monolith]

amy.ames.aims:

[novel cargo]

well there are cases when the limit exists but the slope does not

or is not differentiable

[delicate rivet]

wait how come x^1/3 isn't differentiable?

[willow bear]

it's not differentiable at 0 because the limit up there with f(x) = x^(1/3) and x = 0 works out to +∞

[delicate rivet]

oHHHh thank you sm

[willow bear]

you have a vertical tangent line at 0

[delicate rivet]

gotcha,,, tytyyyyyy

[novel cargo]

like abs(x) at zero is not differentiable but has a limit

[willow bear]

no, $\lim_{h \to 0} \frac{|0+h| - |0|}{h}$ does not exist.

[obsidian monolith]

Ann:

[delicate rivet]

abs(x)?

[novel cargo]

|x|

[delicate rivet]

ohh

[willow bear]

az, what you might have been trying to say is that the absolute value function is continuous but not differentiable at 0

[novel cargo]

yeah,

[willow bear]

which is true, but (a) tangential to what i was saying and (b) poorly phrased on your end

[novel cargo]

ahaha

I'm poorly phrased, baby yoda

what you expect?

[delicate rivet]

does anyone have an example problem of "find the tangen of [x] curve at point [z]?

i need to practice and I've already done all the example problems in my book O-O

[willow bear]

oh sure i can make you some if you want

[delicate rivet]

yes please!

[willow bear]

how many do you want

[novel cargo]

like this?

[delicate rivet]

uh what every suits you, but maybe at least 3?

[willow bear]

ok

[delicate rivet]

wait do you have any problems about finding the equation of the tangent (i don't think finding the gradient of the graph will be on the exam)?

[willow bear]

In each exercise, find the equation of the tangent line to the graph of y = f(x) at x = c.

[1] f(x) = 100x/(x^2 + 1); c = 2
[2] f(x) = cos(x) - 11x; c = 2π
[3] f(x) = e^(5x) + cos(2πx); c = 3
[4] f(x) = sqrt(x)/(x+7); c = 9
[5] f(x) = sin(2πx^2); c = 3/2

[delicate rivet]

ty!

[delicate rivet]

is 1) y=8(4x-3) & 4) (-3x+75)/1696?

[willow bear]

(1) doesn't match my answer key

[delicate rivet]

oh shoot, let me try again

[willow bear]

and neither does (4)

[delicate rivet]

hmm i'll go read over my textbook and try again then

ty for this

is 1) y = 8(2x+1)?

[willow bear]

no

[delicate rivet]

Could you check over my working please?

[willow bear]

f'(x) is wrong

100x is not a constant

[delicate rivet]

oH, ty let me try that again ;-;

um,,, how would you find the derivative of 1) quickly?

[willow bear]

quotient rule

[delicate rivet]

I think my teacher skipped over that? we only learnt the chain rule

could you go over it with me please?

[willow bear]

wait you didn't learn the quotient rule??

ok then you'll have to use product & chain

[delicate rivet]

I googled the quotient rule is this right?

For f’(x)

[willow bear]

very first line, you mis-parenthesized 100(x^2+1)

[delicate rivet]

dont mind the bad handwriting,

[willow bear]

lol

that's bad handwriting?

i've seen way worse

anyway

lemme check this

,w d/dx 100x/(x^2+1)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Ann#0413

[willow bear]

aight yeah you're good

[delicate rivet]

oh yea, sorry about my brackets they're just there so i can write in the multiplication symbol so i remember better

[willow bear]

$100 \times x^2 + 1$ reads as $100x^2 + 1$ and not $100(x^2+1)$

[obsidian monolith]

Ann:

[willow bear]

order of operations

[delicate rivet]

oof yea you're right,,, my bad, i'll fix that

is 1) -4(3x+4)?

[willow bear]

no

[delicate rivet]

Can you show me the proper way on how to do it please?

If youre not busy ofc

[willow bear]

y - 40 = -12(x-2)

[delicate rivet]

do you not then bring the -40 to the other side creating
y = -12(x-2) -40 ?

[willow bear]

no such thing as "bringing" anything to the other side

also, you wouldn't go from z - 6 = 11 to z = 11 - 6 would you?

[delicate rivet]

ok bad wording, do you not add 40 to both sides?

[willow bear]

add 40 yes

but you did not do that

[delicate rivet]

oH skdjhdkj stupid mistake
it then you would expand and factorise for the answer correct?

[willow bear]

you don't really need to factor the 4 out tbh

[delicate rivet]

really? doesn't the answer need to be in the simplest form?

the answer is -12x + 64
or
-4(3x-16) right?

[willow bear]

-12x + 64 is honestly much simpler than -4(3x-16)

it's slope intercept form

[delicate rivet]

ohhh right right

thank you so much for your help,
||i can't believe it took that long for me to do one question 😭 - the amount of careless mistakes i made-||

[willow bear]

yeah the careless algebra mistakes will cost you a lot of time

[delicate rivet]

ikkk it's one of my worst habits,,,,, idk how to fix it tho,,, i just hope i get enough sleep the day before the exam

[willow bear]

do algebra but slowly

[viscid thistle]

how to fix algebra mistakes 101 : do more problems related to simple algebra and a lot of them

[somber folio]

how do i multiply these to be in this form

[willow bear]

you... just do? $\frac{\sin(x)}{x \cos(x)}$ can be broken down into $\frac{\sin(x)}{x} \cdot \frac{1}{\cos(x)}$ just fine

[obsidian monolith]

Ann:

[somber folio]

ohhh okok thanks

[lime bolt]

lol what was confusing about that

[grim sand]

Is G(X) the same thing as F(X)?

[lime bolt]

what is f(x)

[grim sand]

Uhh

F (x) = ?

Do they mean the same thing just different letters or no

[lime bolt]

i dont know what you are on about

[novel cargo]

verified this, which was relatively difficult for me

[grim sand]

I don’t know either anymore

[novel cargo]

oh, sorry, didn't realize conv was going on

[viscid thistle]

f and g are names of functions, you can also call it apple

apple(x)

[grim sand]

So it is the same thing

[lime bolt]

what is the same thing?

[viscid thistle]

yes

[grim sand]

F and G

yes
@viscid thistle thank u

[viscid thistle]

They are both functions

[lime bolt]

well it depends on what f(x)

[grim sand]

So when I solve for F(x) I can also solve for G(x) the same exact way

[lime bolt]

if f(x) isnt g(x) then they arent the same thing

[grim sand]

Right

It doesn’t tell me what F(X) is, only g(x)

[lime bolt]

if and only if f(x)=g(x)

[viscid thistle]

He wasn't sure whether calling a function f or g was the same thing

[lime bolt]

[viscid thistle]

The name of a function is arbitrary

[willow bear]

@grim sand g is just a name

sometimes when you have two or three functions they are called f, g and h

but it's not a hard rule

it's like calling a variable y instead of x

[grim sand]

So when I try to sketch piece wise functions it’s usually F(x) but this equation is G(x), I’m asking if I can sketch it the same way as I sketch F(x) or is there something different

[novel cargo]

the name of the function has no bearing on its graph

the name is arbitrary

the equation that defines the function makes it clear how to plot the graph

also the domain

[viscid thistle]

What if you name your function f(x)=69(x)=5x

[grim sand]

Lost I am

Understood I’m not

[viscid thistle]

Ignore anything I said

(just stop answering the question)

[grim sand]

Same thing or not the same thing

[novel cargo]

are the equations that define G(x) and F(x) the same?

[viscid thistle]

f=G

[novel cargo]

also the domains?

[grim sand]

Yes

[novel cargo]

then they are the same

[grim sand]

Ok

Thanks

[novel cargo]

the could be Z(x) or Barbie(x)

[viscid thistle]

If f(x)=G(x) for any point x then f=G

[novel cargo]

any tips?

[viscid thistle]

Start by rewriting it as tangent

Then rewrite the tangent

[willow bear]

make a right triangle#

seriously make a right triangle

make two of its sides 3 and 8 in a way that gets you an angle with sine 3/8

[novel cargo]

oh, rewrite the inner piece as an algebraic expression?

[viscid thistle]

He can just use $\cot(\arcsin(y))=\frac{\sqrt{1-y²}}{y}$ too

Oh I was going for an algebraic way

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

That works too

[novel cargo]

👍

[willow bear]

bruh ok everyone kind of just went and ignored me

[viscid thistle]

bruh ok everyone kind of just went and ignored me
Your method leads to what Al suggested

Yeah

[viscid thistle]

Yo we have this activity and thw teacher didnt even teach it

How can J find the equation of the ellipes with the center of origin (0,0) the foci have coordinates (4,0) and (-4,0) and a vertex is at (-5,0)

[blissful ridge]

In the general equation of ellipse
$\pm ae=foci$ and vertex is $\pm a$

[obsidian monolith]

The Godfather:

[blissful ridge]

And we have a relation
$b^2=a^2(1-e^2)$

[obsidian monolith]

The Godfather:

[viscid thistle]

Oh its conic section of an ellipse

[mighty onyx]

$ln (1+x^2) + \frac12\ln x - \ln (\sin x)$ would this rewritten be
$((1+x^2)x^1_2/sin(x))$

[obsidian monolith]

Albot1288:

[viscid thistle]

almost.

where did the ln go

[mighty onyx]

he says to not put it in only the numbers after it.

[nova wedge]

question regarding the x and y switch

can someone remind me why the y wouldn't go on top?

is that because we are not dividing by 1/y?

[viscid thistle]

he says to not put it in only the numbers after it.
@mighty onyx what?

@nova wedge taken channel, please move

[nova wedge]

@viscid thistle I'm sorry? Where do I move to?

[mighty onyx]

yeah like it's set up ln A, A= BLANK

or Log A, A= Blank

here is an example

I need to rewrite $ \log_2x + 5\log_2y +-3\log_2z$

[obsidian monolith]

Albot1288:

[mighty onyx]

and the right answer for some reason is $xy^5/z^3$

[obsidian monolith]

Albot1288:

[mighty onyx]

I have no idea why he removed the Parenthesis

I.. I hate it

[viscid thistle]

Like wtf

$\log_2(x)+5\log_2(y)-3\log_2(z)=\log_2(\frac{xy⁵}{z^{3}})$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

And that's it

The log doesn't get removed

And if it really shows (xy⁵)/(z³) as an answer, the book is wrong

@mighty onyx

[mighty onyx]

I'm having a mental breakdown why did he do it like this

Doctorate Holder

[viscid thistle]

Post a pic

If they really did it, they are just wrong

[mighty onyx]

But This one

[viscid thistle]

Ahhh seee

They say: log(A), what is A?

Then they are correct

It is not

[mighty onyx]

I am trying to think of what he put for the fraction exponent

[viscid thistle]

They also even note: ln is not part of your answer

[mighty onyx]

because it should be ((1+x^2)x^1_2/sin(x))

[viscid thistle]

Wdym by _2?

[mighty onyx]

yeah that's what I told yall

to make it $b^1_2$

[obsidian monolith]

Albot1288:

[viscid thistle]

They have x^1_2 as an answer?

[mighty onyx]

how do you do fraction exponents

[viscid thistle]

x^(1/2)

Not x^1/2

Is that where you fell off

Probably yeah

You should have written x^(1/2) and not x^1/2

[mighty onyx]

yep I changed the whole answer to $(1+x^2)x^(1/2)/(sinx)$

[obsidian monolith]

Albot1288:

[viscid thistle]

On latex it is x^{1/2} or x^{\frac12}

[mighty onyx]

I think he did this problem wrong because wouldn't $Log_5(125)=3$

[echo wagon]

It makes a lot of sense that $\frac12$ would work, and yet I never realized it

[obsidian monolith]

Albot1288:

Lunasong:

[viscid thistle]

Yeah

It only works with numbers though Luna

[echo wagon]

Ah, otherwise it looks like a different function

[viscid thistle]

@mighty onyx yes log_5(125)=3

Because 125=5^3

[mighty onyx]

okay so I think he typo'd because all the other ones I put were right but it says it's wrong. What if he put like 2 lol

[viscid thistle]

Wdym by what if he put like 2

#books-old or #book-recommendations for help on books

[mighty onyx]

as a typo idk what he put

know this one has me confused to

Use the Laws of logarithms to rewrite the expression $\ln\left(\frac{x^4\sqrt{x-1}}{3x-14}\right)$ in a form with no logarithm of a product, quotient or power. After rewriting we have $A\ln x+B\ln(x-1)+C\ln(3x-14)$

[obsidian monolith]

Albot1288:

[mighty onyx]

we need to find A,B, and C

[viscid thistle]

Yeah

So basically wants you to use the $$\log(a\cdot b)=\log(a)+\log(b)$$
$$\log(\frac{a}{b})=\log(a)-\log(b)$$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

@mighty onyx

[mighty onyx]

so for B would it turn into (x/1)

he says we need to find the constant for each one

[viscid thistle]

$\log(ab³)=A\log(a)+B\log(b³)=\underbrace{1}{A=1}\log(a)+\underbrace{3}{B=3}\log(b)$

This is an example of how it'd be done

[mighty onyx]

forgot a $

[viscid thistle]

True

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Because you know $\log(a^b)=b\log(a)$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Logarithmic laws

@mighty onyx hello?

Do you need me to explain even further?

[mighty onyx]

I'm still here just trying to work it out now

[viscid thistle]

Okay

Look at the example for reference, tag me when you have a doubt

[mighty onyx]

okay so a=4 b=1/2 and c=-1

[viscid thistle]

Correct. @mighty onyx

[mighty onyx]

[mighty onyx]

would we use the same way for $\log_2\left(32(a+1)^{-4}\right)$ rewritten as

[obsidian monolith]

Albot1288:

[mighty onyx]

$A+B\log_2(C)$

[obsidian monolith]

Albot1288:

[viscid thistle]

Yeah

Why not

[mighty onyx]

for B it's -4

[viscid thistle]

Yes

[mighty onyx]

but what happens to the 32?

[viscid thistle]

Express 32 in terms of 2

And remember that $\log_a(a)=1$

[obsidian monolith]

Al𝟛dium:

[mighty onyx]

32 in terms of 2? like 16

[viscid thistle]

81 in terms of 3 would be 3³

32 in terms of 2? like 16
32 can be expressed in a way where the 2 is the base raised to some number

As i did with 81

[mighty onyx]

oh so it would be for 32 2^5

[viscid thistle]

Yes.

[mighty onyx]

and know c

[viscid thistle]

Do you know how to continue remembering that $\log_a(a)=1$

[obsidian monolith]

Al𝟛dium:

[mighty onyx]

I have these

I have these

but what is your way

[viscid thistle]

After doing 32=2⁵, $\log_2(2⁵)=5\underbrace{\log_2(2)}_{1}=5$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

You can use your logic to know why log_2(2) is 1, think of it this way, to what should you raise 2 to receive 2 as an answer.

@mighty onyx

[mighty onyx]

yeah

[viscid thistle]

All good then?

[olive quartz]

Find the solution of the following equation whose argument is strictly between 180 degrees and 270 degrees.
Round your answer to the nearest thousandth.

$z^5=-243i$

[obsidian monolith]

Rubidinium:

[olive quartz]

I'm not really sure of how I would do this

I think the modulus would be -3 though

as for the angle I have no clue

[hallow thunder]

@olive quartz Hint: $-i = e^{i\cdot 270^{\circ}}$

[obsidian monolith]

jedben2:

[viscid thistle]

Can i get help with deriving powers?

[olive quartz]

what

[hallow thunder]

Do you know euler's formula?

[olive quartz]

$re^{i\theta}$

[obsidian monolith]

Rubidinium:

[hallow thunder]

yeah

[lime bolt]

radians is better than degrees here

[mighty onyx]

@viscid thistle I am still having trouble finding c

[hallow thunder]

yeah but the answer is meant to be in degrees

i think

[olive quartz]

how can we use the formula here?

[mighty onyx]

oops sorry

[hallow thunder]

If you look on the complex plane, $-i$ is $270^{\circ}$ anticlockwise rotation from the center and has a magnitude (length) of 1 from the center. Therefore it can be written as $e^{i \cdot 270^{\circ}}$

[olive quartz]

oh

[obsidian monolith]

jedben2:

[olive quartz]

oh yeah

that makes sense

[hallow thunder]

Ill let you figure out the rest of how to do the question 🙂

[olive quartz]

ok

thanks

[hallow thunder]

np

[viscid thistle]

@viscid thistle I am still having trouble finding c
@mighty onyx repost it

[mighty onyx]

$ \log_2\left(32(a+1)^{-4}\right)$

[obsidian monolith]

Albot1288:

[mighty onyx]

$A+B\log_2(C)$

[obsidian monolith]

Albot1288:

[viscid thistle]

wait you can't find C? it is pretty much the same as the other ones we've done

let's see if it gets clarified with this

remember what we said $\log(a\cdot b)=\log(a)+\log(b)$ right so on this case, a and b are: $$\log_2(\overbrace{32}^{a}\cdot \overbrace{(a-1)^{-4}}^{b})$$

[obsidian monolith]

Al𝟛dium:

[mighty onyx]

so if Log(c) = log(a+b)?

no that's not it

[viscid thistle]

what?

[mighty onyx]

is*

[viscid thistle]

okay

would you be able to express this as a sum of logarithms $\log(4\cdot u^{3})$?

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

@mighty onyx

[mighty onyx]

$\log4 + 3\logu$

[obsidian monolith]

Albot1288:

[stuck lark]

\log

[viscid thistle]

okay

so

[obsidian monolith]

Albot1288:
Compile Error! Click the reaction for details. (You may edit your message)

[stuck lark]

$\log(4)+3\log(u)$

[obsidian monolith]

RokettoJanpu:

[mighty onyx]

yeah thx

[viscid thistle]

$\log(a\cdot b)=\log(a)+\log(b)$ using a substitution with the $u:= a-1$
$$\log_2(32\cdot \overbrace{(a-1)^{-4}}^{u^{-4}})=\log(32\cdot u^{-4})$$ would you be able to express this as a sum of logarithms

[obsidian monolith]

Al𝟛dium:

[mighty onyx]

$\log 32 and -4\log (a-1)$

[obsidian monolith]

Albot1288:

[viscid thistle]

replace the "and" with a sum and you just found C

and well, simplify the log(32) too

[mighty onyx]

I'm still confused I think

how can the constant be the sum of log(32) and -4log(a-1)

[viscid thistle]

which constant

C?

[mighty onyx]

yeah

[viscid thistle]

$A+B\log_2(C)$ doesn't this have a similarity with what you just said

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

you just said correctly $\log_2(32)+ -4\log_2(a-1)$

[obsidian monolith]

Al𝟛dium:

[mighty onyx]

yeah but how can we combine them?

[viscid thistle]

${\color{green}{A}}+{\color{blue}{B}}\log_2({\color{red}{C}})$
$\ {\color{green}{\log_2(32)}}+ {\color{blue}{(-4)}}\log_2({\color{red}{a-1}})$

wdym by combine?

[mighty onyx]

so C would be a-1

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

yes!

[mighty onyx]

fuck my teacher I had tried that like an hour ago but the page says it's wrong

[viscid thistle]

huh

[viscid thistle]

bruh

you told me it was a-1 not a+1

[mighty onyx]

Albot1288:
@obsidian monolith

[viscid thistle]

oh wait did we really skipped a sign

ugh

well that's the typo

[mighty onyx]

so a+1 is the correct answer

I did this one easy tho

[viscid thistle]

yep

good job

[mighty onyx]

thanks

[mighty onyx]

when doing approximation with.
$\log_b(2)=0.271, \log_b(3)=0.429, and \log_b(5)=0.629.$

[obsidian monolith]

Albot1288:

[mighty onyx]

trying to find $\log_b(25b^3)$ should be using power rule right?

[obsidian monolith]

Albot1288:

[viscid thistle]

not only power rules

[mighty onyx]

I know for $log_b(10)$ you break it down into $\log_b(2*5)$ then into $\log_b(2) + \log_b(5)$

[obsidian monolith]

Albot1288:

[mighty onyx]

when using substitution we get 0.9

it's the $b^3$ that's confusing me

[obsidian monolith]

Albot1288:

[mighty onyx]

ooh wait a minute unless we get $25b_3$ by $3\log_b25$ I think

[obsidian monolith]

Albot1288:

[novel cargo]

what is logb(b^3)

[viscid thistle]

For the $\log_b(25b³)=\log_b(25)+\log_b(b³)$ you can use this

[obsidian monolith]

Al𝟛dium:

[mighty onyx]

ok so we would do $(\log_b(5) + \log_b(5)) + \log_b(b^3)$

[obsidian monolith]

Albot1288:

[novel cargo]

you can further simplify if you like

[mighty onyx]

ooh wait log rules $\Log_b(b)^x=x$ right

[obsidian monolith]

Albot1288:
Compile Error! Click the reaction for details. (You may edit your message)

[novel cargo]

you mean logb(b^x) = x

[mighty onyx]

so it would be $0.629 + 0.629 + 3$

[obsidian monolith]

Albot1288:

[mighty onyx]

and yeah

[novel cargo]

would also look better to write 2 * logb(5) + 3

[mighty onyx]

yeah

Yay look at me learning my log rules

[viscid thistle]

Ok i'm back

Oh-

Good job

[novel cargo]

sin(x) + 1 = cos(x)

domain: [0, 2pi)

I got, x = 0, x = 3pi/2

correct?

[stuck lark]

@novel cargo

[mighty onyx]

$10^{2\log5}$

[viscid thistle]

^{}

[obsidian monolith]

Albot1288:

[mighty onyx]

thx, and how would we evaluate this?

[viscid thistle]

is log base 10?

[novel cargo]

2log5 = log(5*10)

[viscid thistle]

?

[proven marten]

yes

[novel cargo]

doesn't help, sorry

[proven marten]

10 is base default

[mighty onyx]

I'm assuming

[viscid thistle]

2log(5)=log(5^2)

[proven marten]

if nothing is written its usually 10

[viscid thistle]

Could also be e

[mighty onyx]

e is for ln

[novel cargo]

I meant 2 log 5 = log(5*2) = log(10) = 1

[proven marten]

doesnt 2 log 5 mean log 5^2

[novel cargo]

oh, man

[viscid thistle]

e is for ln
In higher math log base 10 is so useless that log is usually base e

[proven marten]

Kinda confused

I thought you were in HS

nvm

[viscid thistle]

I am

But I have my resources

[mighty onyx]

I'm in college rip

but back to the problem, would we need to turn it into $\log5^2$ then get that and raise 10 to the power of what we get?

[obsidian monolith]

Albot1288:

[novel cargo]

this simplifies to 25

2 log(5) = log(25)

[proven marten]

agree

[novel cargo]

which is asking what should we raise 10 to to get 25

take 10 to that power

which gives 25

someone write please the algebra and rules

[mighty onyx]

ahhh thanks

[novel cargo]

I can't rememebr them

[viscid thistle]

How about $\log(30)\approx 1/2$

[obsidian monolith]

HoboSas:

[viscid thistle]

$log(25)\approx \log(30)\approx 3/2\approx 1 \approx x \forall x\in \mathbb{N} $

[verbal stump]

log30=1+log3>1>1/2

[lime bolt]

do u mean 3/2

[viscid thistle]

Yes

3/2

[lime bolt]

lol that would have been some serious engineering approximations

[obsidian monolith]

HoboSas:

[viscid thistle]

Cursed

[lime bolt]

i cant see the image

oh ok

[mighty onyx]

$ \ln\left(\frac{a^2}{b^{-4}c^{-1}}\right)=$ expanded is $\ln(a)^2 - \ln(b)^-4 +\ln(c)^-1$ right?

[lime bolt]

what is that green thing

[verbal stump]

$log(25)\approx \log(30)\approx 3/2\approx 2 \approx 3 \approx 4 \approx 10000000000$

[obsidian monolith]

Δφ:

[viscid thistle]

Oxygen tube

Anyway log(25) is close to 3/2

[obsidian monolith]

Albot1288:

[novel cargo]

ln(a^2) - ln(b^4) - ln(c)

[viscid thistle]

Wait base is 10

[novel cargo]

taking the powers out of parenthesis means something else

sorry, ln(a^2) + ln(b^4) + ln(c)

[mighty onyx]

wouldn't be ln(c^-1)

[viscid thistle]

Which is 2ln(|a|) + 4ln(|4|) + ln(c)

[novel cargo]

either -ln(c^-1) or ln(c^1)

from denominator to nominator power changes sign

but it becomes a multiplication and not division anymore

so its subtraction and not addition

[mighty onyx]

so if a=2 b=3 and c=5 then we plug them in to find the answer right

[novel cargo]

I think that is OK

but others may know for sure

[mighty onyx]

I keep getting ln(1620) is that what yall are getting?

[viscid thistle]

Is 1620=2^2x3^4x5?

[mighty onyx]

yes

[viscid thistle]

Correct

[mighty onyx]

so I should be getting 1620? because it says it's wrong

[viscid thistle]

No

ln(1620)

[mighty onyx]

but it says it's still wrong

[viscid thistle]

No way

Show question

[mighty onyx]

and if I'm right the second one should be 30

[proven marten]

Kind of have to ask this

is this for a test

[viscid thistle]

Due in 7 days

I don't think so

[proven marten]

oh

true

ig its just an assignment

[viscid thistle]

Try evaluating it using a calculator

[old nimbus]

When brining a x up from the bottom of the fraction

do you add or times it

[proven marten]

Boris