precalculus

[old nimbus]

I was gonna say that yeah

cheers

[jade heron]

i would put this area

[old nimbus]

yeah

[patent beacon]

Gray, or Grey?

[old nimbus]

Grey?

[uncut mulch]

griy the noone's happy

[blazing raven]

16 - the area between the parabola and the x-axis bounded by x = -2 and x = 2.

32/3 ?

[old nimbus]

you talking to me?

[blazing raven]

It doesn’t matter ....

[old nimbus]

yeah that was the answer

thanks

[blazing raven]

Sorry. I am tired.

[old nimbus]

dont be

if you're asking im not much help

[blazing raven]

Stay up to the screams of the night 😛

[old nimbus]

when integrating (y-1)^1/2

do I use the chain rule or no?

wait nvm thats only for differentiation

[uncut mulch]

generally you should apply chain rule

[old nimbus]

Would I raise that power by 1 and times that wout the front?

[uncut mulch]

are you integrating wrt y?

[old nimbus]

wrt?

[uncut mulch]

with respect to

[old nimbus]

Im not sure

my equation is x=(y-1)^1/2

[patent beacon]

What's your goal?

[old nimbus]

would it be 1/2(y-1)^3/2?

just integrating

oh ok then

nvm I got it

Can anyone explain to me the trapezium rule?

[patent beacon]

Get reimann sums in general?

How an integral can be interpreted as an infinite sum of the areas of rectangles?

[old nimbus]

im not sure

Do you know the equation?

[patent beacon]

Huh?

Equation for what hah?

[old nimbus]

The trapezium rule

[patent beacon]

It's the average of the left riemann sum, and the right riemann sum

[old nimbus]

not sure what that means

could you help me with some problems

[patent beacon]

Feel free to post

[old nimbus]

okay

can you help me with quetion 1?

[blazing raven]

Never trust a politician’s math ...

[old nimbus]

why?

[novel cargo]

you guys learning definite integrals in precalc?

[proven marten]

pog

[novel cargo]

hey JC

where are you?

didn't see you a few days

[proven marten]

Was 'grinding'

Just like you

[novel cargo]

nice

haha

[past meadow]

grinding on your mum

[blazing raven]

I grin ding ..?

[old nimbus]

yeah I am az

why is that a bad thing?

[proven marten]

What textbook is this

[novel cargo]

why is that a bad thing?
@old nimbus I'm not judging. I don't have the knowledge to form an opinion on this matter. I just thought this is more calculus stuff.

[old nimbus]

Oh haha

Sorry

[novel cargo]

generally it isn't a bad thing doing more advanced things

[old nimbus]

Is it advanced?

[novel cargo]

calculus starts with limits, then derivatives, applications of derivatives, anti derivatives/integrals, definite integrals, applicaitons of integrals...

[old nimbus]

Ahhhh

So I'm doing more advanced work then

[novel cargo]

yeah, I would say so

there are layers to all these stuff so

you go over it once like inch deep

next time foot deep

and then you drown

[harsh smelt]

and then you are out or R^3

[old nimbus]

Oh haha

[novel cargo]

^

[harsh smelt]

because you are in R^n

[novel cargo]

commander nailed it

[harsh smelt]

[blissful ridge]

[novel cargo]

what's the benefit of parametric equations?

in linear equations, I see they are separating the movement on the x and y axis into two equations

but what's the benefit?

[willow bear]

have you done physics yet

if you interpret the parameter as time, your parametric equations become the equations describing the motion of a particle

and the curve becomes its trajectory

also this is more of a calculus thing but parameterizations let you calculate the lengths of curves using integrals

[novel cargo]

ahh, so I'm gonna see it later

the real benefit

[harsh smelt]

implicit differentiation also

[novel cargo]

doing a little bit more trigonometry and some more analytic geometry, I deem myself somewhat ready for calculus

really excited about it

hope I don't lose enthusiasm by then lol

looking at it, I see that parametrizaiton enables us to introduce a new independent variable

time, like Ann said

[willow bear]

it doesnt need to be time but it gives a good visual if you interpret it as time

hell it's very often called t

which helps

[novel cargo]

yeah, time feels intuitively an independent variable that can be introduced in almost any equation

I mean even in real life

[novel cargo]

$x(t) = \frac{1}{4}t + 5$

[obsidian monolith]

az:

[novel cargo]

$y(t) = t^2 + t$

[obsidian monolith]

az:

[novel cargo]

is there a way to plot them in wolfram (or other) other then combine them first?

is there a way to combine them with wolfram to check my work?

this give a parabola

$y(x) = 2x^2 - 6x + 5$

[obsidian monolith]

az:

[viscid thistle]

is there a way to plot them in wolfram (or other) other then combine them first?
Wym?

[novel cargo]

right now, to plot the graph, I write it in terms of x and y (combine them) and then plot

isn't there a way to feed the system the parametaized version with t as independent variable and get the plot

and also maybe the combine version?

just to check my work

I'm also new to this subject so maybe I'm expressing my problem wrongly or something idk

[uncut mulch]

just tell it to plot (1/4 t + 5, t^2 + t)

[viscid thistle]

Use desmos

[uncut mulch]

if you're doing it in desmos set an appropriate interval for t

[novel cargo]

can wolfram also combine them for me?

[uncut mulch]

,w plot (1/4 t + 5, t^2 + t)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by ramonov#0921

[willow bear]

@novel cargo desmos does parametric equations!

[novel cargo]

I'm struggling with it right now

haven't still found out

[novel cargo]

lol

that's what I was doing

got it

is there also a way to get the unparametarized equation?

the combined one

oh, man

hyperbola in parametarized form looks really weird

[somber folio]

hi how do i know the values in here

why are the numbers skipping

[near monolith]

What exactly is the question?

[willow bear]

are you asking how to know the coordinates of those grid lines

cause those are 10 and -10

they were not labeled so as to reduce clutter

[somber folio]

yess

how do i know that its 10

[near monolith]

Well they are halfway to the origin

[somber folio]

do i just divide it by 2

this be 9?

[near monolith]

That should work

[somber folio]

can you pls explain how that works D:

[novel cargo]

is this a trig equation?

[willow bear]

the tick marks on each axis are equally spaced

[somber folio]

no i jsut wanna learn how to read these coordinates

[novel cargo]

ok, sorry

[near monolith]

For the closest marked line, you should count how many grid lines up it is

In this case, 18 is up at the 2nd grid line

Then to find out distance for a single gridline, you divide

18 / 2 = 9

[somber folio]

its up 2 times

ohhhh

okay thanks for explaning

[near monolith]

If 18 were on the third one up, then what would the first grid line up be?

[somber folio]

6

pls be 6

[near monolith]

Yes

[viscid thistle]

👍

[viscid thistle]

Need to find the area of the rectangle without the area of the semi circle

Like the smaller parts

Sure

So $\underbrace{A_{\text{rectangle}}}{length \times width}-\underbrace{A{\text{semicircle}}}{πr²}=A{\text{what you want}}$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

What’s the formula for area of a semi circle

Do you know the known formula for the whole circle

A=3.14r^2?

Yeah but use π and not an approximation

Then if that is the whole circle, half the circle what do you think it'll be?

(3.14r^2)/2

Good job but i insist, π not 3.14

Alright

So $\underbrace{A_{\text{rectangle}}}{{\text{length}} \times {\text{width}}}-\underbrace{A{\text{semicircle}}}{πr²}=A{\text{what you want}}$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

@viscid thistle all good now?

I have one more I think

[novel cargo]

aledium, how to you write the pi sign?

[lime bolt]

copy and paste it

or have like an extension which has symbols

[novel cargo]

oh

[novel cargo]

are these fractions?

[lime bolt]

notice that 9-x^2 can be factored as DOTS

which cancels the 3-x

[viscid thistle]

Dots?

[lime bolt]

difference of 2 squares

[novel cargo]

multiplication

[lime bolt]

im assuming u mean the two upper and lower chunks are meant to be bracketed right?

as this has multiple interpretations

[viscid thistle]

This is a simplification I believe

Yeah

[lime bolt]

so cancel 3-x first

then probably times both chunks by 3+x

[viscid thistle]

aledium, how to you write the pi sign?
I have it in my keyboard, if not use \pi

[novel cargo]

how on keyboard?

special keyboard, hardware-wise?

[stuck lark]

greek keyboard or hotkeys

[viscid thistle]

I don't have greek letters tho

Only pi and delta

I'm mobile

[lime bolt]

oh yea on mobile you can get custom keyboard extensions

but then it is also possible on pc

[novel cargo]

true

[somber folio]

am i interpreting the graph correctly

[viscid thistle]

Those limits are correct indeed

Are you drawing the graph?

[novel cargo]

just wanted to say the graph could be for this limits (although I don't know limits still very well)

[somber folio]

nope these are pre drawn choices from khan academy

thank youu

[viscid thistle]

Np

[opaque olive]

$f(x)=x^m(x-1)^n\f'(x)=x^m(x-1)^n(mx^{-1}+n(x-1)^{-1})$

[obsidian monolith]

Yes:

[opaque olive]

how would i find the second derivative from f(x)

[stuck lark]

f''=(f')'

[opaque olive]

shall i just use product rule again

or is there a simplification to make it easier?

[stuck lark]

whatever makes sense

[viscid thistle]

yo

is swokowski a good text book?

[opaque olive]

his calculus book?

[viscid thistle]

precalc

alg with trig and analytical geometry

[weak needle]

Haven't tried it.

His geometry and trigonometry is pretty good, though

[opaque olive]

$(mx^{-2}+n(x-1)^{-2})+a(mx^{-1}+n(x-1)^{-1})+b(mx^{-1}+n(x-1)^{-1})$

[obsidian monolith]

Yes:

[opaque olive]

can i even factorise that anymore

cuz i need to show $\frac{m}{n+m}$ is negative

[obsidian monolith]

Yes:

[opaque olive]

https://gyazo.com/742aa74e7d2bf69da8f957f89d6b62fd

ive done the first part ...

this is the part concerning the second derivative

<@&286206848099549185>

[blazing raven]

Er ... so what do you have so far?

Do you have a formula for the x-value at the stationary point?

Take two derivatives and verify that that point is a minimum / maximum by finding all possible x values and showing the x from the first part is "the one."

[novel cargo]

for a I have: $33\frac{1}{3} \cdot 2\pi = \frac{22\pi_{rad}}{min}$

for b I have: $22\pi \cdot 12 = \frac{264\pi_{in}}{min}$

[obsidian monolith]

az:

az:

[novel cargo]

can someone check?

thanks

[willow bear]

33 1/3 here does not mean 33*1/3

it's a mixed fraction

33 + 1/3

[novel cargo]

oof

so 200pi/3

and 24000pi

2400pi

[willow bear]

200pi/3 radians per minute yeah

and no it should be 800pi inches per minute for the second part

wait no

400pi

the radius is 6 in

[novel cargo]

oh man

my brain is on strike today

thanks

[novel cargo]

why is it so important to remove irational numbers/radicals from denominator?

[willow bear]

it's not, but usually radicals in the numerator are considered simpler than radicals in the denominator

that's subject to context tho

[novel cargo]

instead of 1/sqrt(2) writing sqrt(2)/2 seems more complicated

[modest mist]

Nah

[novel cargo]

using Qalculate!

really like it

[cedar cipher]

why is it so important to remove irational numbers/radicals from denominator?
@novel cargo makes it simpler to approximate

1/√2 would be 1/(1.414..) let's say you stop at 3 places. 1/1.414 is more tedious than calculating 1.414/2

[novel cargo]

true, makes sense

[novel cargo]

$\csc{\theta} = 3$

[obsidian monolith]

az:

[novel cargo]

$\sec(90^{\circ} - \theta) = \sec(\frac{\pi}{2} - \theta) = \csc \theta = 3$

[obsidian monolith]

az:

[novel cargo]

someone check please

[willow bear]

well so far you haven't done anything wrong but also haven't said much either

nor have you been clear what exactly your goal is here

are you trying to solve this equation for θ?

[novel cargo]

was an exercise that asked to find other trig functions based on the given one using identities

csc theta was given

and sec(90 - theta) was sought

but I can solve it for theta too

good idea

[willow bear]

...

why did i mention this

[novel cargo]

because you are much more advanced than I am

and think in your domain

your level

which is normal

I understand and appreciate that

learning a lot

[willow bear]

i mean ok like

alright

you're given csc(θ) and asked for sec(90°-θ) and you've recognized that those are in fact one and the same

so good job

[novel cargo]

so we have 1/ sin theta = 3 or sin theta = 1/3

looking up sin theta = 1/3 we get 0.34 rad

sin's period is 2pi

so we have 0.34 + n * 2pi for n being all integers including zero

just not sure if I'm missing half of the solutions

[willow bear]

you are

pi - arcsin(1/3) is another solution

[novel cargo]

so we also have (pi - 0.34) + n * 2pi

shouldn't a calculator give me both solutions when I input arcsin(1/3)?

[willow bear]

no

arcsin isn't multivalued

arcsin(x) is the angle between -pi/2 and +pi/2 whose sine is x

and that's unambiguous and refers to one angle and one angle only

[novel cargo]

so the fourth quadrant and the first

so, I have to imagine the solutions in the four quadrants, think about the graph, its parity and stuff and make an educated calculation

[novel cargo]

may be a stupid question, but what's the benefit/use of reference angles?

[uncut mulch]

easily shift between quadrants

[novel cargo]

what do you mean?

[uncut mulch]

sry vague wording,
makes it relatively easy to express general solutions
can also help evaluate stuff like sin(240°) without a calculator / memorising every single value on the unit circle

[novel cargo]

when I see something greater than 2pi i subtract the greatest possible multiple of 2pi from it and have a standard angle

or add, if it's negative

I think I'm missing something obvious here

[uncut mulch]

what would you do with something like sin(240°) though (without access to a diagram of the unit circle, calc or memorising all of it)

[novel cargo]

I'd say it's in the third quadrand

negative for both sin and cos

also, equal to 240 - 60

sign filpped for sin

like thinking in the quadrants and moving the terminal side and imagining what happens with the x and y values

[uncut mulch]

also, equal to 240 - 60
not quite

[novel cargo]

oh sorry

[uncut mulch]

the reference angle would be 240° - 180° = 60°

[novel cargo]

yes

240 - 60 - 60 I meant

[uncut mulch]

why subtract 60 twice?

[novel cargo]

subtract 60 once to get to the horizontal (180) line and another time to move back the same amount to have the same x and y values

[uncut mulch]

that's inefficient

[novel cargo]

agree

[uncut mulch]

and can be inconsistent

[novel cargo]

yeah

you have to reason about it every time

and prone to error

[uncut mulch]

the reference angle is the acute angle made between the terminal arm and the x-axis

[novel cargo]

oh man

that's what I'm doing all along just made in a formal and consistent way

the reference angle I mean

[uncut mulch]

in this case, that can be determined by subtracting 180° from 240°

[novel cargo]

true

[uncut mulch]

as indicated in the middle diagram

[novel cargo]

the same angle just in the first quadrant

[uncut mulch]

hence since 240° is in quadrant 3
sin(240°) = -sin(60°)

[novel cargo]

you are turning on some lights in my head, good sir

[uncut mulch]

would be an application of reference angles

angles don't need to necessarily be nice, and you can apply the same idea to get a decent idea of the size

[novel cargo]

angles being nice?

[uncut mulch]

non-special

[novel cargo]

ahh

yeah, some of them are really naughty

It's starting to make sense

thanks Ramonov

[rapid falcon]

There are two wheels with a belt around both of them. If I'm trying to find the linear speed of the belt (v=radius(omega)), then do I add up the radii and omega of both circles before multiplying?... or do I just use the values of the larger circle?

[novel cargo]

you have to calculate the angular speed of one of the wheels in rad then multiply that with that wheel's radius

[rapid falcon]

thanks

[mighty onyx]

Anyone good at expontential functions?

[lime bolt]

yes

[viscid thistle]

@mighty onyx just ask it

[fossil crown]

From

#❓how-to-get-help

Rule 1

The help channels are solely for help with math, so feel free to post your question. Asking whether you can ask a question or if anyone knows about some specific topic is unnecessary, so please try to avoid questions of that nature.

[mighty onyx]

Give a formula for the function illustrated using a vertical shift of an exponential function. The two points marked on the graph are A=(−1,−7) and B=(1,2). The red horizontal line is given by y=5, and is a horizontal asymptote of the function.

[viscid thistle]

can we see the function too? @mighty onyx

[mighty onyx]

That's all it gives me besides Y=_____

[viscid thistle]

it seems like you were given a illustration

[viscid thistle]

okay so $f(x)=a(b)^x+c$ we'll use this, a marks the y-intercept as $(0, a)$, b represents the growth and c represents the horizontal asymptote

[obsidian monolith]

Al𝟛dium:

[mighty onyx]

okay

[viscid thistle]

so you are given 2 points, but we already know that c=5 in our case, so you can make a system of eqn for the rest

[mighty onyx]

But what would it be?

I got two different answers but they are both wrong

[viscid thistle]

can i see them?

with the work you did to get them

this channel is occupied, i'll tag you on a free one so that i can help you.

[dusk solstice]

ok

[viscid thistle]

We have a parabola based on number of sales over a long period of time where s = number of sales and t = time (in months). The equation representing the number of sales as a function of time is: s = -0.7t2 + 8t + 6
In excel, create a t-chart for this function.
In Cell A7 label t (time) In Cell B7 label s = -0.7t2 + 8t + 6
In cells A8 and A9 you get to choose the t value we wish. Choose 2 values for t.
Using excel, find the average rate of change between those 2 points.
Chose another 2 values for t and find the average rate of change between those two points.

could someone help me out here with this question?

this channel is still occupied, i'll tag you on a free one so that i can help you.

sorry

[weak needle]

Is not that just Excel?

[viscid thistle]

yeah but i have no knowledge on excel

[mighty onyx]

@viscid thistle

Here is my work the 1.069×1.8708^1 works but not for 7 and the other one is just wrong

And don't worry about the log problem

[opaque olive]

$(a+b)^{2}\leq2a^2+2b^2$

[obsidian monolith]

Yes:

[opaque olive]

how do i show this is true?

im guessing i start with the fact that (a+b)^2 is bigger or equal to zero

but it doesnt get me anywhere

[viscid thistle]

Here is my work the 1.069×1.8708^1 works but not for 7 and the other one is just wrong
@mighty onyx do you have the solutions?

Also i don't understand your letter

[mighty onyx]

I don't have the solutions I just went on to my other homework dealing with logarithms. I will ask my teacher what is the answer.

[uncut mulch]

@opaque olive 2a^2+2b^2 = (a+b)^2 + (a-b)^2

[opaque olive]

yah got it

thanks

[opaque olive]

https://gyazo.com/a155bf1c66c3bb028f3246735ac7c1e1

im having difficulty calculating the height

of the pyramid

could anyone help

[steady seal]

how do i go about this
https://i.imgur.com/ajdrXCS.png

[uncut mulch]

get rid of skyscraper fractions

[steady seal]

o this is pre calc channel lol

and ya I figured but forgot how to do it lmao

[uncut mulch]

multiply and divide by lcd of relevant fractions

in this case: 2(2+x)

[steady seal]

o right

ty

I kept multiying by 2+x only for some reason

[tame wedge]

does this type of function have a name?

Has form e^(at)*sin(sqrt(1-a^2)t)

[jade heron]

If a<0 u could call it damped sinusoidal I think

[novel cargo]

oh man

I'm starting to understand some of the stuff that wolfram throws at me

happy

[harsh smelt]

,w lambert-W function

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Commander Vimes#7532

[novel cargo]

what function is W even?

[stuck lark]

google lambert W

[novel cargo]

multivalued function

sounds like advance calculus

[stuck lark]

not really

[novel cargo]

what then?

[hallow thunder]

it is the inverse function of $xe^x$ @novel cargo

[obsidian monolith]

jedben2:

[novel cargo]

thanks

[somber folio]

Am i evaluating this correctly

[harsh smelt]

looks like yes

,w lim sqrt((x^2-3x-1)/(x^3+1)) as to -1

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Commander Vimes#7532

[harsh smelt]

hm

ah

-4

[somber folio]

oh did i do something wrong

[harsh smelt]

,w lim sqrt((x^2-3x-4)/(x^3+1)) as to -1

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Commander Vimes#7532

[harsh smelt]

nah, you are correct

[somber folio]

thank youu

[somber folio]

Am i graphing this correctly

[viscid thistle]

Uh no

[harsh smelt]

branch 4-x is sus

it has negative slope

but as graphed, positive

and second branch

why it is 1?

[somber folio]

i thought it was like y=mx+b

[viscid thistle]

x+4 is also a bit sus

[somber folio]

y intercept would be 4 in x+4 and slope would be 1/1 right

[harsh smelt]

@somber folio your slope looks right

but what is x+4 at x = -4?

[somber folio]

okok leme try and think for it a bit

isnt that bcuz of the implications

[rustic mortar]

does this type of function have a name?
@tame wedge
This is the bessel function if i'm not wrong

one of the bessel functions atleast

[viscid thistle]

It is a piecewise function. @somber folio

[somber folio]

hm ok leme just try again

[novel cargo]

cos(x - pi/2) is cos(x) shifted to the right by pi/2 which equals sin(x)

what is cost(pi/2 - x)?

I can't wrap my head around this

[viscid thistle]

cost?

[hallow thunder]

*cos

[viscid thistle]

@hallow thunder cos(pi/2-x) is just sin(x) too

[hallow thunder]

yeah ik

its just when he put cost(pi/2 - x) it was a typo

[novel cargo]

I admit

but how is cost(pi/2 -x) also sin(x)?

isn't it moving horizontally to the left?

[willow bear]

going from x to pi/2 - x is a reflection across the vertical line x=pi/4

[viscid thistle]

because cos(x)=cos(-x)

[hallow thunder]

since $\cos(x)$ is an even function, we can factor out $-1$ from the inside: $\cos(\frac{\pi}{2} - x) = \cos(-( x - \frac{\pi}{2})) = \cos(x-\frac{\pi}{2}) = \sin(x)$

[obsidian monolith]

jedben2:

[novel cargo]

oh, you are making too much sense, good sir

thanks

still trying to understand Ann's argument

it sounds very enlightening

[willow bear]

reflecting across the line x=a gets you from y = f(x) to y = f(2a-x)

[novel cargo]

this is just true with trig functions because they periodic, r?

[hallow thunder]

(sorry for crappy writing)

[novel cargo]

wow

I feel good

[willow bear]

periodicity doesn't matter here

[novel cargo]

true

I'm starting to see

also, a side question:

how much do I have to be good with precalc before starting calc?

this is an open ended question, IK

I'm asking bc there is so much to it

[hallow thunder]

i would have a basic understanding of derivatives, integrals and limits.

[novel cargo]

I have

especially the concepts

[hallow thunder]

thats good then

[novel cargo]

not the techniques necessarily

derivatives give you the rate of change

the tangent line

the slope of the graph at a particular point

the notion comes from limits

integrals are the inverse or something of derivatives

they give you the area under the graph (I really don't understand this one lol)

[slow roost]

lol this is for pre calc review but its more of like alg: if f(x)=4x-x^2 find f(d)-f(-4)

[willow bear]

what's giving you trouble here claire

[slow roost]

can someone check if i got it right or if i made a mistake somewhere? i got 32

ik how to do it, its just i dont have the answer key unlocked

so i cant rlly check

lol i think im prone to mistakes due to the - signs

[hallow thunder]

@novel cargo i would also have a look at finding derivatives of some functions too since you will do complicated ones in calc

[novel cargo]

the answer can't be some simple number

@slow roost

[slow roost]

what?

it shpuld be tho

lol

its just solving functions

[novel cargo]

maybe I'm wrong

[slow roost]

oops f(4)-f(-4)

[willow bear]

f(d) - f(-4)?

[novel cargo]

bc f(d)

[slow roost]

mb i mistyped it

[willow bear]

,w f(x) = 4x-x^2; calculate f(4) - f(-4)

[obsidian monolith]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by Ann#0413

[willow bear]

wtf

ok desmos time

[novel cargo]

got it @hallow thunder

[willow bear]

yup 32 checks out

[slow roost]

yay

tysm

what did u use for desmos by chance

i dont have access to our answer key yet

[slow roost]

cause the modules are locked until 10am . my time sigh

is it the graphing calc or

[willow bear]

yes

[novel cargo]

didn't know this worked

[slow roost]

ah tysm!!!

how about if sqrt f(3/2)

i tried desmos and my answer was off by decimals

i got 2sqrt3-3/2

not sure if there was a calculation error

[heady lotus]

this should be the answer imo

Since sqrt is not distributive over addition or subtraction

You can't take the square root of the numbers individually and add/subtract them

[viscid thistle]

Yeah $\sqrt{a-b}\neq \sqrt{a}-\sqrt{b}$

[obsidian monolith]

Al𝟛dium:

[slow roost]

yeah im not sure

[viscid thistle]

About what?

Do you want to know how to simplify correctly $\sqrt{6-\frac94}$?

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Tho there is not much of simplification

[slow roost]

sure

[viscid thistle]

Ok know how to do common denominator?

[slow roost]

mhm

[viscid thistle]

How should i interpret that?

[slow roost]

for 6-9/4 under the sqrt?

[viscid thistle]

Yes

[slow roost]

multiply the 6 by 4 to get 24/4

[viscid thistle]

Yes

[slow roost]

so thats 24/4-9/4= 15/4 well sqrt 15/4

[viscid thistle]

Yes

[slow roost]

how did u guys get the sqrt6-9/4 tho

[viscid thistle]

And then as $\sqrt{\frac{m}{n}}=\frac{\sqrt{m}}{\sqrt{n}}$ you can simplify a little bit further

[obsidian monolith]

Al𝟛dium:

[slow roost]

how am i suppose to plug it in properly tho

cause i did 4(sqrt3/2) - (-sqrt3/2)^ hence why i got 2sqrt3-3/2

but clearly thats incorrect

[viscid thistle]

$f(x)=4x-x²$ would you be able to get f(3/2)?

[obsidian monolith]

Al𝟛dium:

[slow roost]

um what?

[viscid thistle]

cause i did 4(sqrt3/2) - (-sqrt3/2)^ hence why i got 2sqrt3-3/2
I see what you did now

[slow roost]

yeah

lol

[viscid thistle]

$\sqrt{f(\frac32)}\neq f(\sqrt{\frac32})$

[slow roost]

oooo

[obsidian monolith]

Al𝟛dium:

[slow roost]

oo

[viscid thistle]

Better

[slow roost]

OOOO

OHHHHHH

wait wait so its

[lime bolt]

bruh

[slow roost]

sqrt (4*3/2)?

hence why its sqrt 6?

LOL

[viscid thistle]

Wait where are you now on

[slow roost]

im on the 4x part

putting sqrt f(3/2) into 4x

lol can u do it on paper or something

from what im understanding this is what yall did

[viscid thistle]

hence why its sqrt 6?

$\sqrt{\frac{4\cdot 3}{2}}=\sqrt{\frac{{\color{green}{4}}\cdot 3}{{\color{green}{2}}}}=\sqrt{{\color{green}{2}}\cdot 3}=\sqrt{6}$

[lime bolt]

it is trivial

[hushed sorrel]

where did ur root go

at the end lol

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Okay let me just write everything out

[slow roost]

lol yeah i understand how u got sqrt6

@hushed sorrel wdym?

[hushed sorrel]

no there was just a typo before i think

ignore me x

[lime bolt]

your pfp 🤔

[hushed sorrel]

its an alpacatrain

do u like it xxx

[slow roost]

lol im pretty sure i understand it now

i just interpreted the subsitution in correctly

i was suppose to plug 3/2 into the 4x and then take the sqrt of it

but i saw it as 4*sqrt3/2

which is correct

its suppose to be sqrt 3*4/2

and then for the -x^2 its (3/2)^2 and then take sqrt of that

so thats sqrt 9/4

[viscid thistle]

$f(x)=4x-x²$ $\sqrt{f(3/2)}$ let's first get f(3/2) and then sqrt the result. $$f(3/2)=4(\frac32)-(\frac32)²=6-\frac94$$ then the sqrt as it is asking for $\sqrt{f(3/2)}$ $$\sqrt{f(3/2)}=\sqrt{6-\frac94}=\sqrt{\frac{15}{4}}=\frac{\sqrt{15}}{2}$$

[slow roost]

mhm yeah i understand it now

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Glad to hear

[slow roost]

tysm!

[somber folio]

am i defining this correctly

[hushed sorrel]

yes

[slow roost]

is it crossed out bc theres no solution or?

[lime bolt]

why would there not be a solution for it

it would just be 2 times the rest

[hushed sorrel]

maybe ur teacher just doesnt want u to do that question idk

[mighty onyx]

Give a formula for the function illustrated using a vertical shift of an exponential function. The two points marked on the graph are A=(−1,−7) and B=(1,2). The red horizontal line is given by y=5, and is a horizontal asymptote of the function.

anyone know how to do this?

[willow bear]

do you know the general form of an exponential function?

[mighty onyx]

Y=a(b)^x

[willow bear]

lowercase y, but yes.

in your case it'll be $y = a \cdot b^x + 5$ due to the vertical shift.

[obsidian monolith]

Ann:

[willow bear]

you have two unknowns to solve for, and you have two (x,y) pairs.

use the given points on the graph to construct two equations in a and b

then solve those equations

[mighty onyx]

so after we enter our two points is should look like $2 = a \cdot b^1 +5$ and $-7 = a \cdot b^2 +5$

[obsidian monolith]

Albot1288:

[mighty onyx]

** $-7 = a \cdot b^-1 +5$

[obsidian monolith]

Albot1288:

[mighty onyx]

right?

[obsidian monolith]

Yes:

[willow bear]

@mighty onyx late reply and ^{-1} but yes

[willow stream]

$\tan(\pi /3)/(\pi/3)^2$?

[obsidian monolith]

Dozenford:

[willow stream]

How would I go about simplifying this

I know tan would resolve to $(sqrt(3)/2)/(1/2))

$(sqrt(3)/2)/(1/2))

[novel cargo]

enclose it with $

[obsidian monolith]

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

[willow stream]

There, I know tan would result in this and the bottom is just pi square over nine

but where do I go from there?

[uncut mulch]

misplaced parentheses in your Tex.
tan(pi/3) would simplify to just sqrt(3)

and then multiply the numerator and denominator by the same thing to get rid of skyscraper fractions

[willow stream]

Oh alright

so before I evaluated tan(pi/3) I would just cancel out pi/3 from top and bottom

[uncut mulch]

no

[willow stream]

so tan (pi/3) is sqrt(3) after you simplify got it

misinterpreted mb

[uncut mulch]

you should also know that without having to write and divide sin(pi/3) by cos(pi/3)

[willow stream]

Yeah that's true

[mighty onyx]

Ok so I got 1/6 for B

[uncut mulch]

you were getting somewhere before, but this is completely wrong

especially since there are square roots of negatives

$\begin{cases} -7 = ab^{-1} + 5
\ 2 = ab^{1} +5
\end{cases}$

[obsidian monolith]

ramonov:

[uncut mulch]

this is the system you should be solving

[mighty onyx]

I got A as +-6 and B as +- 1/2

[uncut mulch]

for b^x to be well defined for all x, b needs to be positive

and as a result there is only one possible value for a here

[mighty onyx]

Here is the work I got using substitution

[uncut mulch]

i can't really read what's going on there

[mighty onyx]

so with the first line I turn my ab^-1 into a/b

next with 2=ab=+5 I turn it into b=-3/a

then -7= a/3 +5

and got +- 6

[uncut mulch]

lots of missing signs in your work

making that substitution would result in
-7 = -a^2/3 + 5
which does lead to a = +-6

however one of the solutions are extraneous for the initial premise

specifically the domain of (standard) exponential functions is all real numbers,
and to satisfy that, the base of the exponent must be positive

eg (-1/2)^(1/2) would be undefined

[mighty onyx]

okay so my teacher just sent me a document but it's crazy

$-6(1/2)^x +5$

[obsidian monolith]

Albot1288:

[uncut mulch]

i.e. using
a = 6 would get you b=-1/2
which is a solution to the system being set up but does not model an exponential function with the given properties.
a = -6 would get you b=1/2
which does work

[mighty onyx]

What's dumb is I had put that in but he says to put the exponent inside the parenthesis so.... rip

Thanks everyone for helping with this hard problem

[uncut mulch]

that's unecessary though

they'd be equivalent

they themselves had the exponent outside

[mighty onyx]

yeah but atleast it's over now

[novel cargo]

I've learned arcsec's domain is -1 <= X <= 1

my calc tho gives:

how's this defined?

gives this for sqrt(-2) as input to arcsec

sorry, was so confused forgot to write that

I interpreted arcsec as 1/arccos

we have to make an exception for zero

bruh, spit it out

[stuck lark]

arcsec ISN'T 1/arccos

[novel cargo]

ahah

I'm out of my depth

going into unknown lands

[echo wagon]

Can your calculator tell me what arcsec(0) is? Since 0 is in your domain

Also, I've never seen a calculator with arcsec on it lol

[novel cargo]

proud of my gangsta qalculatar!

not a quitter

[echo wagon]

arcsec(-sqrt(2)) is not the same as arcsec(sqrt(-2)) btw

[novel cargo]

I'm sorry, I mistyped

but, really good catch Luna

attention to detail

[echo wagon]

Is this a troll post lol?

[novel cargo]

the original question?

[echo wagon]

Yes

[novel cargo]

I wish I were that advanced

lol

nah, I just learned new things

and am exploring

[echo wagon]

Okay, but why is your calculator giving you that value? Lol

[novel cargo]

my calc is for real

[echo wagon]

,w calculate arcsec(-sqrt(2))

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Lunasong#7197

[echo wagon]

This is the right answer lol

[novel cargo]

https://qalculate.github.io/

this is my calculator

[echo wagon]

Stop using it

[novel cargo]

you have any suggestions? Ubuntu

[stuck lark]

https://cdn.discordapp.com/attachments/363224154469826562/747553561382224050/unknown.png what's this trash

[novel cargo]

my calculator

[echo wagon]

At least arcsec(0) isn't defined, so they made up a value

[novel cargo]

I take full responsibility

[echo wagon]

But arcsec(-sqrt(2)) is actually defined and they just give some other value

[novel cargo]

lol

[stuck lark]

imagine using an rng to guess the evaluation at pts outside domain of defn

[echo wagon]

Lmao

[novel cargo]

I don't fully understand but sounds funny to me too

[echo wagon]

What do you think and is it a test?

[narrow pier]

anyone know what this is all ab

its IXL

its practice

[stuck lark]

eg evaluating $x\mapsto\frac1x$ at 0, but instead of saying undefined the calc says $\frac\pi e$

[obsidian monolith]

RokettoJanpu:

[narrow pier]

wow

eg evaluating $x\mapsto\frac1x$ at 0, but instead of saying undefined the calc says $\frac\pi e$

[obsidian monolith]

Turnip:

[novel cargo]

Turnip learns fast

[narrow pier]

hahahahahha

[stuck lark]

$\tan(\frac\pi2)=\phi^{-\sqrt2}$

[obsidian monolith]

RokettoJanpu:

[narrow pier]

u know what i gotta learn tho

that

[echo wagon]

@novel cargo BTW the domain of arcsec is $(-\infty,-1] \cup [1,\infty)$, so literally the opposite of what you had

[viscid thistle]

what is "this function"

[narrow pier]

its impossible

[obsidian monolith]

Lunasong:

[narrow pier]

@viscid thistle ^^

[novel cargo]

it really depends on context

[echo wagon]

They probably mean the set of those points

[narrow pier]

yea

[viscid thistle]

oh, I thought those were multiple choice options

[echo wagon]

Anyway, what do you think the range is Turnip?

[stuck lark]

arcsec is defined as the inverse of sec where the domain of sec is restricted to [0,pi]\{pi/2}

[narrow pier]

difference betweeen highest and lowest?

and just list them out?

[echo wagon]

Define range of a function

This is not like the range of a data set

[narrow pier]

its the difference between the highest and lowest in a set

[echo wagon]

No

[narrow pier]

yikes

[stuck lark]

$\mathrm{im}(f):=\brc{f(x):x\in\mathrm{dom}(f)}$

[obsidian monolith]

RokettoJanpu:

[echo wagon]

An informal description would be: The range of a function is the set of all output values of a function

[stuck lark]

the set of f's images as you vary f's input over f's domain

[novel cargo]

when Turnip is done, explain to me the difference between image and range

[stuck lark]

diff name same thing

[narrow pier]

An informal description would be: The range of a function is the set of all output values of a function

knowing that what is a step i could take to solving

[lime bolt]

you havent showed the fucntion yet

[echo wagon]

Look at all the output values, put them in a set

They have

[novel cargo]

knowing that what is a step i could take to solving
@narrow pier consider all possible outputs for the possible inputs. that's the range

[echo wagon]

It's ${(17,17), (14,6),(2,-6),(13,17) } $

[obsidian monolith]

Lunasong:

[narrow pier]

do i just put them together>

?

[stuck lark]

@novel cargo *codomain

[novel cargo]

OruniJetak, I once read a linear algebra book and they made a difference between the two

[lime bolt]

no that is still the range

[novel cargo]

maybe it was the codomain

[echo wagon]

Can you tell us what the outputs are for the function? @narrow pier

[lime bolt]

every element of the codomain doesnt have to be an output

[stuck lark]

no diff @novel cargo

[lime bolt]

there is a difference

[stuck lark]

not talking to you

[echo wagon]

I've seen the codomain called the range too, but that's dumb and should be avoided.

[narrow pier]

17 -> 17
14 -> 6
2 -> -6
13 -> 17

[echo wagon]

So the outputs are?

[stuck lark]

range is far more often synonymous with image

[lime bolt]

it is equivalent

[narrow pier]

what do you mean outputs

[stuck lark]

diff name same thing

[lime bolt]

what the function produces given an input

[echo wagon]

If x is in the input, f(x) is the output

@novel cargo look like they disappeared. So where did you get this whack calculator?

[novel cargo]

just a second

[lime bolt]

lol getting those calculators with graphs and stuff is pretty silly

[novel cargo]

@narrow pier

[echo wagon]

Lol

[lime bolt]

i think my school is gonna make me pay like £80 for one

[novel cargo]

all made with GNU software too

[lime bolt]

but im just gonna say no

[novel cargo]

you can't put rocks in there

and you won't get gold out of it

[stuck lark]

it's a rock grinder

[uncut mulch]

hand-held graphics calcs should be banned

[novel cargo]

@echo wagon I searched for good calcs on Ubuntu and one of the good according to reviews was Qalculate!

https://qalculate.github.io/

[echo wagon]

I think to get students to rely less on their calculators, they all need to be given az's calculator

[novel cargo]

no one recommended it

[uncut mulch]

tables

[novel cargo]

I fee actually cool with this qalculator

feel

lol

[lime bolt]

is a high-functioning caclulator necessary

[novel cargo]

no, but a high energy one for sure

like mine

[echo wagon]

No, I literally use an abacus

[novel cargo]

https://cdn.discordapp.com/attachments/363224154469826562/747553561382224050/unknown.png

look what it does

[stuck lark]

🤢

[novel cargo]

the bestest of calculators

[echo wagon]

LOL

I didn't even notice the i/0 in the answer

[stuck lark]

make it compute $\sqrt[i]{i}$

[obsidian monolith]

RokettoJanpu:

[novel cargo]

high energy

lol

[stuck lark]

with usual branch cut

[novel cargo]

says degree must be rational

[echo wagon]

So it can do i/0 but not that? Lame

[novel cargo]

yeah, I'm disapointed

[stuck lark]

lame

[low owl]

college said "buy a graphing calculator" -- so I buy one....College Algebra last semester : You can't use that. --- Precal this semester : You can't use that

don't buy graphing calculators lol

[velvet blade]

Why to buy them? Can't we have a small monitor and keyboard setup with desmos in it while being tested?

[delicate rivet]

what does it mean by "signed area"?

i googled it, and it highlighted the area under the curve between certain points, but is that not the area / distance travelled?

[low owl]

Why to buy them? Can't we have a small monitor and keyboard setup with desmos in it while being tested?
They strictly enforce a "nothing except pen/pencil, paper, and a scientific calculator" policy here by making you take your exams via Lockdown Browser + Live Webcam. Have to show your work-area lol

[cursive rock]

hey im new here

[viscid thistle]

@delicate rivet can you explain yourself better? I'm not getting what you are trying to say, the area below the function on a velocity time graph is indeed the distance traveled

[cursive rock]

Why does 1 over X2+X = 1 over x2 + x3 over x2? I dont understand

[uncut mulch]

what's written on the page is incomplete but

$\frac{1}{x^2} + x = \frac{1}{x^2} + x \cdot \frac{x^2}{x^2} = \frac{1}{x^2} + \frac{x^3}{x^2}$

[obsidian monolith]

ramonov:

[delicate rivet]

@delicate rivet can you explain yourself better? I'm not getting what you are trying to say, the area below the function on a velocity time graph is indeed the distance traveled
@viscid thistle in the picture i sent the first dot point said that "the signed area between a v-t graph and horizonal axis is the displacement x", could you please explain what signed area is?

here's the picture

[bright ravine]

if the area is below the x-axis, then it is considered negative (i.e. in this case it would mean negative displacement)

[novel cargo]

$sec[arcsin(x-1)]$

[obsidian monolith]

az:

[novel cargo]

$arcsin(x-1) = u$

[obsidian monolith]

az:

[novel cargo]

$\sin{ u} = x - 1$

[obsidian monolith]

az:

[harsh smelt]

\

[novel cargo]

$\cos{u} = \sqrt{1 - (x-1)^2}$

[obsidian monolith]

az:

[novel cargo]

$\sec{u} = \sec{[\arcsin{(x-1)}]} = \frac{1}{\sqrt{1 - (x-1)^2}}$

[obsidian monolith]

az:

[novel cargo]

$0 < x \leq 1$

[obsidian monolith]

az:

[novel cargo]

can someone please check

if this is correct

[past meadow]

,w sec(arcsin(x-1))

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Sneaky#0928

[novel cargo]

oh my

[past meadow]

it is poggers

[novel cargo]

thank you

I feel dumb

but I've learned this just now

feel also proud

[past meadow]

pretty poggers tbh

[novel cargo]

lol

[past meadow]

couldnt you be less restrictive with the domain tho

[novel cargo]

yeah, I'm looking at wolfram

and they've 0 to 2

[past meadow]

like you could have $0<x\leq 2$

yeah

[obsidian monolith]

Sneaky:

[past meadow]

because

domain of arcsin is $[-1,1]$

[obsidian monolith]

Sneaky:

[past meadow]

so all you care about is $(x-1)$ being in $[-1,1]$

[obsidian monolith]

Sneaky:

[past meadow]

(and x!=0)

those conditions give $0<x\leq 2$

[obsidian monolith]

Sneaky:

[past meadow]

actually

you dont want x=2 either

[novel cargo]

yeah, bc zero

[past meadow]

so yeah, $0<x<2$

[obsidian monolith]

Sneaky:

[novel cargo]

👍

[severe prawn]

A bus company collected a total of $1080 from
passengers one morning. Each passenger paid either
a $3.00 regular fare or a $1.50 reduced fair. If 7 of
every 8 passengers paid the regular fare, what is the
total amount of money, in dollars, that the bus
company collected from passengers paying the
regular fare?

[novel cargo]

you can convert this maybe to a two equations two unknowns system

[severe prawn]

doesnt work for me

[novel cargo]

how about this

1080 = 3x + 1.5y

1080 = 7/8(x+y) * 3 + 1/8(x+y) * 1.5

[severe prawn]

check on wolfram

what does it get for x?

[novel cargo]

do it yourself

important is you understand the equations

[severe prawn]

i do

is x 1008

[novel cargo]

no

can you explain to me what the equations mean?

[severe prawn]

but the correct answer is 1008

[novel cargo]

what does 1008 stand for?

[severe prawn]

x

[novel cargo]

what does x stand for?

[severe prawn]

number of ppl payed regular fare

3 dollors

[novel cargo]

so, you saying that 1008 ppl paid 3 dollars in addition to others who paid less would result to 1080, which is the total amount collected?

do I understand you correctly?

[severe prawn]

no

1008 is 3x

[novel cargo]

so then what is x?

[severe prawn]

okay

got it

[novel cargo]

look at the equations, and try to describe them in plain english, is my advice

[severe prawn]

i understand them i just had a brain fart

with this problem

ty

[novel cargo]

then OK

np

[viscid thistle]

w as s u p

[novel cargo]

hey

[viscid thistle]

👋

Ahh yes, spending 45 minutes on a problem only to factor and then unfactor

[novel cargo]

haha

[viscid thistle]

Thought I was getting close to the solution and babam I ended up where I started

Turns out

I made the silliest mistake on it

And did an oops with a factor

I did something like (3x-2)(x^2-1)(x+6) and the (x+6) was supposed to be subtracted

Anyways

What y’all up to?

[novel cargo]

I'm doing trig

solved tan[arccos(x/2)]

[viscid thistle]

Beautiful

without calculator?

without calculator?

[novel cargo]

yes

[viscid thistle]

How’d you do it

[novel cargo]

arccos(x/2) = u

[viscid thistle]

Mhm

[novel cargo]

so cos u = x/2

[viscid thistle]

Mhm

[novel cargo]

now you can say adj/hyp = x/2

[viscid thistle]

m h m

I think there’s another way to get cos(u) = x/2

but continue

[novel cargo]

from there you can calc sin u = sqrt(4 - x^2)

[viscid thistle]

m h m

[novel cargo]

now having both cos and sin

sin u / cos u = tan u

[viscid thistle]

Agreed

[novel cargo]

this in my precalc book but marked as calc

[viscid thistle]

mhm

[delicate rivet]

to find the min & max, you find the stationary right? I solved for the derivative, but I got plus/minus root6/3.... how would you do this by and is this even correct?

[blissful ridge]

Do you mean $\frac{\sqrt{6}}{3}$

[obsidian monolith]

The Godfather:

[willow bear]

you need to compare the values of f at:

• the stationary points inside your domain
• the boundary points of your domain

so in this case, f(0), f(2) and f(sqrt(6)/3)

the smallest among these will be your min and the largest your max

[delicate rivet]

thank you!

[delicate rivet]

how did they get 1.9?

i subbed in 0, 2 & ±sqrt6/3 and got:
(0, 3); (2, 7); (sqrt6/3, (4 x sqrt6)/6 + 3); (-s/qrt6/3, (-8 x sqrt6)/6 + 3);

[uncut mulch]

seems they did an approximation

you don't need to calculate f(-sqrt(6)/3) since -sqrt(6)/3 isn't in your domain

[delicate rivet]

oooo right right, but i put it into the calculator and it didn't come out as 1.9?

[uncut mulch]

did you put appropriate parentheses?

[austere void]

-sqrt(6)/3 is roughly -0.8

[delicate rivet]

yep, im p sure i did; i put in (4*sqrt6)/9 + 3

it came out as 4.08 which would round up to 4.1

[uncut mulch]

how are you getting (4*sqrt6)/9 + 3

[delicate rivet]

since the stationary point is positive/negative sqrt 2/3
so i subbed it back into the first equation

(sqrt(2/3))^3 - 2*sqrt(2/3) + 3

[uncut mulch]

check your signs