precalculus

[storm stirrup]

Drawing

[viscid thistle]

Alright nice

Got 10.1

Thank you based bird

[storm stirrup]

10.1 is a yes from me

[viscid thistle]

Let's get out of here or wha? 😏

[storm stirrup]

S-sen...

...pai

U want to....

[viscid thistle]

Um I gotta finish my homework 😳

[storm stirrup]

[real fog]

do i differentiate this or integrate can someone pls help

[viscid thistle]

Integrate

[real fog]

but why

[prime prawn]

integrate

cuz it says to

[viscid thistle]

Because it tells you to integrate lol

=Tex \int

[prime prawn]

$$/int$$ represents an integral

[viscid thistle]

Rip

[real fog]

is nt Sf(x) the integrated version

[prime prawn]

$$\int$$ represents an integral

no.

[prime prawn]

Do: $$\int (x+1/x)^2dx$$

[real fog]

y= f (x) is the interagrated one right

[prime prawn]

for part a

no.

[real fog]

what is it

[prime prawn]

you are doing $$\int f(x)$$

[real fog]

thats the difrentiated version

[prime prawn]

no, $$f(x)$$ is the differentiated version

[real fog]

so from f(x) to Sf(x) it gets intgrated right

[prime prawn]

Part A: $$\int f(x)dx=\int \left(x+\frac 1x \right)^2dx$$

[real fog]

i know how to integrate but i dont undestand the question it ask

[prime prawn]

The question asks you to integrate f(x)

depending on what f(x) is

[real fog]

f(x) is it the questions (b)(a)

https://cdn.discordapp.com/attachments/363224154469826562/443022652588687360/unknown.png

cause it says when f(x) is given

@prime prawn

[prime prawn]

wrong post

[real fog]

yes i get it but ist f(x) the integrated version

[prime prawn]

It says when "$$f(x)$$ is given", "find $$\int f(x)dx$$"

[real fog]

so when the diffrentiated version is giiven i have to find the integrated one right

[prime prawn]

yes.

[real fog]

but i have other questions here which say find Sf(x)

and the i integrated them

*then

[prime prawn]

When it says that, to find $$f(x)$$ given $$\int f(x)dx$$, you differentiate $$\int f(x)dx$$ to find $$f(x)$$

[real fog]

loo at the q on the right

8look

*look

[prime prawn]

numberr 7 right

[real fog]

4 5 and 6 it says integrate

[prime prawn]

oh 4 5 6

You find the integrl fr number 4,5 and 6

[real fog]

yea

[prime prawn]

but not for number 7

[real fog]

yea but you said its already integrated

Sf(x) is alrady integrated

[prime prawn]

I need help for this one question. Verify $$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\csc A+\cot A$$

[viscid thistle]

<@&286206848099549185> I'm kinda shaky on this one

It's a toughie, solved in #geometry-and-trigonometry

[hexed ermine]

@prime prawn For verifying trig identifies, the best advice is to evaluate everything in sine and cosine, notice how the RHS can be simplified into them

So you want to work with the RHS

[prime prawn]

@hexed ermine doesn't always work

[simple vault]

I have a ti-84 calculator but I don't know how to use it with this problem

[viscid thistle]

Go to STAT

Click edit

and input those values

Next go to Sin regression

And it’ll give you the sinusoidal equation of best fit

[simple vault]

how do I go to sin regression? @viscid thistle

[viscid thistle]

go to STAT

go to calculate

and at the very bottom there should be sin regression

@simple vault

[simple vault]

what period do I use? @viscid thistle

[viscid thistle]

what do you mean what period? you mean for the screen of the graph?

[simple vault]

I clicked on sinregression

and it wants me to input a period

[simple vault]

@viscid thistle

[viscid thistle]

when i press sin reg it automatically calculates it :/

try putting in something like 1 for the period

im using a ti-83 so it might be slightly different

[simple vault]

http://prntscr.com/jfl207

dunno how this is wrong

[dire crypt]

try substituting in 210 degrees

[viscid thistle]

“Find all degree solutions”

You need a solution set like 10+10z

[wide frost]

remember an angle+360° is (at least for sin and cos) the "same" angle

[simple vault]

http://prntscr.com/jfysms

<@&286206848099549185> I cant figure this one out

[sand dock]

Turn it into a quadratic equation first

Let $$x=\sin{2\theta}$$

[sand dock]

$$x^2-6x-1=0$$

[sand dock]

This needs quadratic formula

$$x=\frac{6\pm\sqrt{36+4}}{2}$$

[simple vault]

(3 +- rad 40)/2?

[sand dock]

Yep

Where you can pull out a 4 and make it a 2

So $$x=\frac{6\pm2\sqrt{10}}{2}=3\pm\sqrt{10}$$

[sand dock]

So now you have two different solutions

[simple vault]

3 + rad10 & 3 -rad10

[sand dock]

$$\begin{array}{lcl}\sin{(2\theta)}=3+\sqrt{10} \ \sin{(2\theta)}=3-\sqrt{10}\end{array}$$

[sand dock]

So then use arc sin to solve for all solutions to each

[simple vault]

I got an error

[sand dock]

You only put one answer though

There should be multiple

[simple vault]

Error for the 3 + rad10

when I do inverse sine

[sand dock]

Trying to look

[rustic nebula]

Its cus 3 plus root 10 is greater than one

Sin x cant be greater than one or less than -1

[rocky bison]

It can

unless real numbers

but im just being a masochist so ignore me

[rustic nebula]

:3 ur right

[viscid thistle]

@rocky bison u're a honorable masochist

[hexed ermine]

@rocky bison gj on Honorable :v

[rocky bison]

Thanks 😃

@viscid thistle The masochistic part will never leave me

[viscid thistle]

@rocky bisonbeing masochist in math isn't bad

That's actually good

[rocky bison]

Maybe

But masochists are insufferable

Or overly masochistic people I should add

[viscid thistle]

Why would you like to suffer in maths?

[rocky bison]

What

[stable horizon]

no one likes to suffer in math

you just do

cause math always finds away to trip you up

just like coding

programming*

[viscid thistle]

Lol the difference is that if you get something wrong in maths or programming, you know that you are definitely wrong.

[swift heath]

does anyone know how to combine two trigonomic wave forms with the same angular frequency but different horizontal displacement?

[true vigil]

use sum of angles formula

https://math.stackexchange.com/questions/535600/sum-of-sinusoids-with-same-frequency-sinusoid-proof

[swift heath]

i dont understand that at all

[true vigil]

well it's a very thorough answer

might want to try to understand it

[swift heath]

it looks years above my current material

[true vigil]

Asin(ωt+ϕ)=Asin(ωt)cos(ϕ)+Acos(ωt)sin(ϕ)=A′sin(ωt)+A′′cos(ωt).

this is the main point

[swift heath]

whats the quote mark for?

[true vigil]

they're another variable

you can call them B and C or whatever instead

so the idea is using the sum of angles formula

if you have Asin(ax + b) and Bsin(ax + c)

then you express both as sums of sin(ax) and cos(bx)

using sum of angles

and then combine them back again

[swift heath]

how do i do that?

[wide frost]

best way to derive are complex exponentials

[true vigil]

sin(a+b) = sin(a)cos(b) + sin(b)cos(a)

it's much easier with complex exponentials?

[swift heath]

oooh

[wide frost]

imo yes

[swift heath]

i need to make it into the format sin(a)cos(b) + sin(b)cos(a)

k

[true vigil]

you're right

sadpone if you know complex polar notation then this is much faster

[wide frost]

see, thats why complex numbers are so beautiful

[swift heath]

i shouldnt know complex numbers at this stage, this is week 5 material, complex numbers are week 7+

normally we use this

[true vigil]

yeah

that's the plan

that's the method that the link I sent you explains

[swift heath]

but not only is it always formatted as sin or cos(nt), its never different numbers or nt+a

[true vigil]

only the notation is a bit concise

what do you mean

[true vigil]

the process is similar

[swift heath]

same angular frequency, never a phase shift

[true vigil]

it doesn't get much more difficult

review this process closely and then read the link again

[swift heath]

i have no precedent for how to handle a phase shift

[true vigil]

it's a very similar idea

it's written in a terse, concise way

so it's a bit hard to read

but it's not really much above your level

[exotic anvil]

Can anyone help with desmos

]

the 3rd equation cancels out the second one for some reason

[viscid thistle]

I think it’s cus you put it as r<1

and r = 1 + [something]

Try putting a , @exotic anvil

a comma

[exotic anvil]

Hmmm

[exotic anvil]

interesting

I got it done anyhow

This took a while

[thick raptor]

lol

[still yew]

Can we see the equations?

[thick raptor]

no

[vague timber]

Ok

[elfin night]

Kk

[sweet steeple]

How do I find the limit as x -> 1 when the equation is 1/(x-1)

[thick raptor]

x-1 gets arbitrarily small

1/(x-1) gets arbitrarily large

limit doesn't exist

[sweet steeple]

what about 1/x when x approaches infinity then?

is that just something i have to remember

[thick raptor]

x gets arbitrarily large

1/x gets arbitrarily small

it approaches 0

[sweet steeple]

ohhh

is that the same with negative infinity then

[thick raptor]

yes

[dense zealot]

Yay u found a way to realize 1/0 dne

[lucid vortex]

Is there a fourmula to convert this into degree and minuites

[hexed ermine]

It'

s just a conversion

there are 60 minutes in 1 degree

so just set a proportion

$$\frac{45}{60}=\frac{x}{1}$$

[hexed ermine]

==45/60

[granite stirrup]

0.75

[hexed ermine]

so Angle B converts to 77.75 degrees

[viscid thistle]

Sorry

Can anyone help me with this

I missed class yesterday cuz of AP stats test

Now the teacher wants me to do this

And im so confused

[exotic briar]

@viscid thistle We can barely read the questions

[viscid thistle]

🤔

Let me try again

Sorry

Hmmm

How is it now

@exotic briar

[exotic briar]

Please clarify

Why are there 4 idential figure 1s?

identical*

[viscid thistle]

i think the teacher wants me to do it in different ways

she was not at school today either

bruh

[wind igloo]

You only go S and E.

You have to go E 5x

[exotic briar]

well I mean it seems like this tbh

https://betterexplained.com/articles/navigate-a-grid-using-combinations-and-permutations/

[wind igloo]

And S 3 times

[exotic briar]

8!/ (5! * 3!)

[wind igloo]

Way to jump right to the answer. Bravo.

[exotic briar]

8 is traveling unit and 5 and 3 are x and y units respectively

http://puu.sh/AnCRB/f9b9600c0f.jpg

[viscid thistle]

hmmm

[exotic briar]

The link I shared

Take a look at those

I think it will explain in details

ask if ur stuck on specific steps

Why there are 4 boxes, I donno

because there is clearly more than 4 possible routes

[viscid thistle]

ye

[exotic briar]

unless I misread the question

[viscid thistle]

there are many different routes

[exotic briar]

@wind igloo I'll fight you AI. U can't talk to a human that way 😛

You know the old saying, look both ways before crossing the street but when you don't... then... really don't kek

[viscid thistle]

i think i did something wrong

she said only 8 blocks long

but somehow i got only 6 blocks, wondering if i did not understand the question well

or even5 blocks

...

[wind igloo]

What did you do?

[viscid thistle]

I just tried to draw and connect them

It seems wrong

[exotic briar]

take pic and show

[viscid thistle]

It cant be this simple right

[wind igloo]

1. I think boxes are houses.

You're supposed to walk along the street.

[viscid thistle]

O

[wind igloo]

And 2) You're not allowed to go diagonally.

[viscid thistle]

Only south and east

How about this

One sec

What does backtracking mean

Like u cant go back?

[wind igloo]

Yeah.

You only go south or east.

You never go north again, or west.

[viscid thistle]

Like this?

Right side is east

[wind igloo]

Right.

[viscid thistle]

Down is south

Ah

[wind igloo]

That's a good route.

Start at the H go to the W.

[exotic briar]

kek

go through the houses xD

I'm dying lol

All this time, google map was messing with me

[viscid thistle]

[exotic briar]

Why take 23rd Street turn when u can just drive through the police station?

Fucking googlee

and their fucking algo

[viscid thistle]

I got it now

Its much easier than i thought

Thank you so much

@exotic briar @wind igloo

[wind igloo]

yw

[exotic briar]

make sure u try other questions likee that

so that u can remember how to do it for other similar questions

Don't be like me

[viscid thistle]

Roger that

Thanks

[exotic briar]

best of luck. I'm gonna make fun of u elsewhere 😛

[prime prawn]

nice profile

how do you find the circumference of an e;lipse

pi r sqrt((a^2+b^2)/2) right?

but where does the sqrt((a^2+b^2)/2) come fro

[sage herald]

actuallu there r no accurate formula for an eclipse only rough approximates

but not sure bout sqrt((a^2+b^2)/2)

[elfin night]

Googles formula they use is very complicated

But yea Bamboo is correct there is no exact formula

[prime prawn]

wait but how do you get the sqrt(average of b^2 and c^2)

[analog raven]

Hello i've been working on a pre calc study guide and i've been stuck on the last few problems for the last two hours can anyone help

[fringe stream]

[analog raven]

can you help me solve cos8θ + cos10θ

[fringe stream]

What do you need to do with it?

[analog raven]

The question says write each product as a sum or a difference with positive arguments

[blazing raven]

well the problem isn’t even a product ...

Are you sure you are giving us the full instructions?

[long pond]

This should probably go in #math-discussion or #discussion

If you want the helper tag, you can inquire with Jacobian I think.

[brave void]

<@&286206848099549185> So I know this is an arithmatic sequence that takes away 19 bales each time

How to evaluate it with the total number of trips, I do not know

[rocky bison]

He doesn't take away 19 each time

ut decreases by 19 each time

Or was that what you meant

[brave void]

Yeah

I mean it decreases by 19 each time

Could I write this as

216 (-19n)

I don't know how to count the hay he takes away for the number trips

It's summation

Or something

[rocky bison]

So he first takes 262 bales

[brave void]

Yes

There's an equation i think

[rocky bison]

So that's 262-0*19

[brave void]

Would I do

Sum of all terms = (n/2)(first term + last term)

So n would be 14

So the sum would be

[rocky bison]

Sum is given by

[brave void]

7(262-4)

[rocky bison]

=tex S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)

[brave void]

what is a

[rocky bison]

a is first term

d is the change between terms

so our sum is like

=tex a+(a+d)+(a+2d)+\dots + (a+(n-1)d)

[rocky bison]

Yes?

So a=262

d=-19

[brave void]

Yes

[rocky bison]

now we're given n too

[brave void]

yes

14

[rocky bison]

So we just plug it all in

and we get an answer

Yes?

[brave void]

1036?

[rocky bison]

Looks good

[brave void]

oh rlly

[rocky bison]

You can always add it all up manually

If you want to be really sure

Just to confirm yourself

I wouldn't suggest it all the time but it's useful to install some confidence in your answers

[brave void]

It says it is incorrect

[rocky bison]

I got 1939

=tex \frac{14}{2}\left(2(262)+(14-1)(-19)\right)

[brave void]

1 sec

could you help me with the last one

then I'm done

It's hard

I've got no idea

[rocky bison]

Ok

Start off

Draw your square

[brave void]

k 1 sec

[rocky bison]

The first one

And label the sides

[brave void]

done

[rocky bison]

What's the area of that square

[brave void]

100

[rocky bison]

Ok

So let's say

A_1=100

Now the second square

If the lines are drawn from the mid points

Then what's the length of each side

(You're gonna need some trig for it)

[brave void]

root 50

pls confirm that lol

[rocky bison]

yeah

I wanna simplify it though

so 5root2

[brave void]

k

[rocky bison]

So if there's a square

with lengths of that

What's the area

[brave void]

50

[rocky bison]

yep

Now we can do one more to be sure

So A_2=50

Now what's A_3

Same idea

Find the lengths of the sides

It'll be half of 5root2

Then find the new square's sides

[brave void]

l

1 sec

[rocky bison]

yep

[brave void]

4/25 as the length of one side?

[rocky bison]

seems a bit off

=pup eval \sqrt{2\left(\frac{5\sqrt2}{2}\right)^2}

[granite stirrup]

Query made by @rocky bison
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=eval+\sqrt{2\left(\frac{5\sqrt2}{2}\right)^2}
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[brave void]

1 sec

[rocky bison]

oke

[brave void]

Wait

Why multiply by 2

[rocky bison]

Just a short cut

It's that + itself

[brave void]

oh yeah

[rocky bison]

Found your error?

[brave void]

1 sec

no

oh wait

found it

[rocky bison]

awesome

ready to continue?

[brave void]

Yes

[rocky bison]

Ok

SO we have a new square

with lengths of 5

What's the area

[brave void]

25

[rocky bison]

awesome

So now we have

A_3=25

If we list them all

100,50,25

Notice anything about them?

[brave void]

1/2

[rocky bison]

Yep

It's a geometric sequence

[brave void]

Yeah

[rocky bison]

where r=1/2

So we want to calculate the sum to infinity of it

=tex \sum_{n=0}^\infty\left(100*\left(\frac{1}{2}\right)^n\right)

[rocky bison]

So do you know the equation for a sum to infinity?

[brave void]

No

sry xD

[rocky bison]

Ok

It's

=tex \frac{a}{1-r}{|r|<0}

[rocky bison]

That domain is kinda important

Because we can't take the infinite sum

If it's bigger than 1

should be 1 not 0

whoops

Because if it's bigger than 1 then the terms keep getting bigger

meaning it just goes to infinity

It's what is called divergent

So we can calculate this

a is first term

[brave void]

Yes

[rocky bison]

r is ratio (What you multiply by between terms)

[brave void]

100/.5

[rocky bison]

yep

[brave void]

It goes to

200

[rocky bison]

yep

That's what I got too 😃

[brave void]

converges

[rocky bison]

Exactly

[brave void]

That's the sum

Of an infinite series

Sequence*]

[rocky bison]

yep

series

[brave void]

It isn't a decimal?

[rocky bison]

sequence is abunch of numbers

like

1,2,3,4,5,6,

series

is sum

1+2+3+4+5+6+7+

[brave void]

They say Sequence in the question

[rocky bison]

Yeah

[brave void]

That's why I was curious

[rocky bison]

Then they ask for the sum

[brave void]

So just

plain

200?

[rocky bison]

Yep

Nice trick huh?

[brave void]

yeah

well that's all I got

thx masn

[rocky bison]

np 😃

Was fun

[warm surge]

can anybody help me with a question?

If a function f is increasing on (a, b) and decreasing on (b, c), then what can be said about the local extremum of f on (a, c)?

[olive briar]

It's at b

@warm surge

And it's probably a max

Think about it draw a line that's increasing from ab

then at b it starts decreasing

[gray kindle]

probably?

wouldn't it HAVE to be a max?

[livid tendon]

@olive briar @warm surge @gray kindle Careful, if a function f increasing on an interval X means f(b) >= f(a) for all b > a where a,b in X (and symmetrically, decreasing just means f(b) <= f(a) for all b > a where a,b in X) then the function f(x) = 1 satisfies both of these definitions on the entire real line, but would you argue that this function has either local or global extrema?

See: https://math.stackexchange.com/questions/1704338/is-a-horizontal-line-an-increasing-or-decreasing-function

[warm surge]

@olive briar thanks

[olive briar]

@gray kindle ^ I was also thinking of a case where b didn't exist

[sage herald]

for a geometric series 1 + r + r ^2 + r ^3 + .. + r ^n-1

how do you find the right formula?

[long pond]

Do you just mean r^n?

[sage herald]

oh yes sorry

[long pond]

Np

[brave void]

<@&286206848099549185> Wouldn't it be -3y+8ny?

[elfin night]

no @brave void

=tex a_n=a_1+\left(n-1\right)d

[brave void]

Okay

So

-3y+7ny?

[elfin night]

Not quite

=tex a_n=-3y+\left(n-1\right)8y

[brave void]

-3y+8ny-8y

-11y+8ny

[elfin night]

Yup

[brave void]

So

[elfin night]

So no. 5

[brave void]

Yeah

And the last one:

So I find the value of the first term

That one is 88

And the other one is

88+(13)(-4)

I think?

Not positive

[brave void]

Okay

So Ssub 14

=

7(88+?)

[elfin night]

Yup

You can find d

Then use that to find the 14th term

[brave void]

But

What is the equation

Is it

88 + (13)(-4)

[elfin night]

Sorry

Gotta tag me or I wont notice lmao

[brave void]

oh xd

[elfin night]

The equation for the 14th term?

[brave void]

yeah

[elfin night]

Yea that looks correct

Then sub that value in for an

=tex S_n=\frac{14}{2}\left(88+88+\left(-4\right)\left(13\right)\right)

I think

[elfin night]

I got 868

[brave void]

7(88+36)

Same

[elfin night]

Should be good then ; )

[brave void]

Thx man!

[viscid thistle]

why is definite integral defined as the area under the curve?

[fading token]

No it's not

Only if some specific conditions hold, the definite integral of a function corresponds to some area under the graph of the function

[brave void]

@elfin night

[elfin night]

=tex \left(-1\right)^n\cdot4^{\left(1-n\right)}

[elfin night]

Is the same as

=tex 4\left(-\frac{1}{4}\right)^n

[elfin night]

=tex \sum_{n=0}^{ \infty}\left(-1\right)^n\cdot4^{\left(1-n\right)}

[brave void]

wait a sec

so I plug in 1

oh wait nvm

I shut up

[elfin night]

; )

=tex S_\infty=\frac{a}{1-r}

[elfin night]

You probs know that one

From geometric sums

[brave void]

yeah

so

[elfin night]

We have a given to us

[brave void]

4 is the first term

[elfin night]

Yup

[brave void]

then 1 + 1/4

Since 1/4 is negative

[elfin night]

Yea

[brave void]

16/5?

[elfin night]

Yup, should be 3.2

[brave void]

cool

thx man

would that be the same

For this one?

I just do that equation

[elfin night]

You'd need a different formula for that one

Since you're not summing to infinity

[brave void]

Is that one

yeah

Is the ratio

(1+ 17/1000)

[elfin night]

Looks it yea

[brave void]

@elfin night

[elfin night]

I think it'll be 20 not 19

Since the first term is 1

So then each successive multiplication was gonna be another term

[brave void]

OOooooh

so that to the twentieth

[elfin night]

Yup

[brave void]

Anyway to compute on math bot?

[elfin night]

I can try

[granite stirrup]

Runtime error in iterm_1
On line 1 at position 1

pup 1+1
^
Failed to access variable pup

[elfin night]

==(1-(1.017)^20)/(1-1.017)

[granite stirrup]

23584615352269932528428138747069394243960844178184775027153/1000000000000000000000000000000000000000000000000000000000 = 23.5846153522699

[brave void]

woah

Yep

Thx man

gtg for memorial day

[elfin night]

Have fun

[mortal basin]

I have a limits questions based on concept

Select the best option that would complete the statement below: A limit of a function fails to exist when.... *

the function is constant.
the graph of the function oscillates.
both one-sided limits of the function equal to each other.
the function passes the continuity test.

is is when the function passes the continuity test

[fading token]

Nah

[mortal basin]

then what is it?

[fading token]

Which one do you guess?

[mortal basin]

the oscilating one

because if they function is constant the limit is the same everywhere

[fading token]

That's correct

Well done c:

[mortal basin]

but why

or maybe im not thinking about it correctly

because that would be like a sine wave right

[fading token]

Correct, but if you for example take that sine wave, then you can exploit its waviness to cook up a very ugly function that doesn't have a limit everywhere

=pup plot sin(1/x)

[granite stirrup]

Query made by @fading token
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=plot+sin(1%2Fx)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[fading token]

this beast doesn't have a limit at 0 because of its very wild oscillations

All the other mentioned properties actually show that the function is well-behaved and has existing limits

[mortal basin]

oh okay i see

thank you very much

[fading token]

🍮

[long pond]

“The graph of the function oscillates” makes me think of sin(x), which wouldn’t make sense because it’s defined for all real numbers

I think that should be a bit more specific

[fading token]

The goal of the question is to give "the best fitting" answer though, which isn't quite precise or rigorous

[long pond]

True

[eternal solstice]

Hey guys. How do i find the horizontal asymptote of #26

I know the vertical asymptote is -1/2 by setting the denominator equal to 0.

I know the answer is 1/2 but I'm not sure how to get to that answer

[tall granite]

=tex \frac{2x}{x-6} = \frac{2}{1 - \frac{6}{x}}

[tall granite]

maybe use this?

and ummm

vertical asymptote is -1/2?

u sure?

youre probably talking about #25

[eternal solstice]

Shoot i totally mean 25

Yeah so for 25 I get -1/2 as the vertical asymptote, but I dont understand how to find the horizontal.

[tall granite]

do the same thing that i did with #26

[eternal solstice]

Ok I think that worked.

Was that just the inverse?

[tall granite]

ummm

idk

[eternal solstice]

ok thanks

appreciate it

[torn nebula]

anyone can help me with trig?

[pine wyvern]

What do you want to know?

[eternal solstice]

Whats up

[tall granite]

yo whadup

[torn nebula]

solve 4cosx +3(sinx)^2-2=0

no idea how to do this

[young fractal]

@torn nebula use the identity

[torn nebula]

I got x=cos^-1((2+sqrt7)/3)

but it says domain error or somethong

what does that mean

[young fractal]

What?

No

@torn nebula you know how to use an identity right

After that, you factor it

[torn nebula]

oh ok thank u

I got cosx(4-3cosx)+1=0

how do I know whether to change to cosine or sine

oo nvm the only identity is sinsquared

[young fractal]

Let me try to do that problem

4cosx+3(1-cosx^2)-2

4cosx+3-3cosx^2-2

-3cosx^2+4cosx+1

3cosx^2-4cosx-1

Let me condense it to x so it’s an easy read after factor

What’s this...

I don’t know where I’m ending up, I’m sure you did yours differently

[jovial sentinel]

hi guys, does anyone happen to know how to do this problem?

[hexed ermine]

Do you know the length of your transverse and longitudinal axis in this situation?

[hollow zenith]

[torn nebula]

can someone help me with number 2

[hexed ermine]

In QI, all three trig functions are positive

[viscid thistle]

omg

there are people who actually talk on this server

[hexed ermine]

To memorize the trig functions, you want to picture the unit circle in your head

Lol, not much going on at this time

[viscid thistle]

oh yeah the unit circle

the one with radians

[hexed ermine]

?

[viscid thistle]

I suppose he's not wrong

[sweet niche]

@torn nebula If you plan on doing physics/math or anything else like that remembering the unit circle is the best thing to do. It'll help you out tremendously if you do it now.

Just watch some vids on what the meaning is before you memorize it though

[hollow zenith]

what is precalculus?

[tranquil stone]

Calculus is the mathematical study of continuous change, in the same way that geometry is the study of shape and algebra is the study of generalizations of arithmetic operations.
Precalculus is a course that includes algebra and trigonometry at a level which is designed to prepare students for the study of calculus. Schools often distinguish between algebra and trigonometry as two separate parts of the coursework.

@hollow zenith

[viscid thistle]

@hollow zenith its an american thing

[hollow zenith]

@tranquil stone Thx

[hollow zenith]

Hey, someone can be my guide on Calculus?

[rocky bison]

[steep fox]

Ok

[digital rock]

right, so you have a triangle with hypotenuse 34, and other sides y and x right

[steep fox]

uhh sure

[digital rock]

but you also have that y = 1.1x

[steep fox]

Yes

[digital rock]

so the right triangle has sides 34, 1.1x and x

where 34 is the hypotenuse

[steep fox]

That is the gradient.

y = 1.1x

The gradient.

This could be plotted on a graph.

That was the original intent.

But I’m curious to see where you were going with it.

[digital rock]

use pythagoras on that right triangle

x^2 + (1.1x)^2 = 34^2

[steep fox]

Yes

[digital rock]

221/100 x^2 = 1156

[steep fox]

However this has already turned into a longer method

[digital rock]

I mean if I were actually doing it x^2 + (1.1x)^2 = 34^2 would be the first line :^)

[steep fox]

Exactly

And since the answer isn’t there

You can already tell it’s going to be a longer method

[hollow zenith]

2nd degree equations?

@digital rock

[steep fox]

Since my answer is one line

[digital rock]

the question is what, to find the area of the rectangle?

[steep fox]

I don’t mean to be rude mudkip

Yes

[digital rock]

I would be interested to see your solution then

[steep fox]

But I did ask if there was a shorter way. One consisting of less calculations is what I meant

I showed you it

(34 x sin(tan^-1(1.1))) x (34 x cos(tan^-1(1.1)))

[digital rock]

(34 x sin(tan^-1(1.1))) x (34 x cos(tan^-1(1.1))) is not the simplified answer

[steep fox]

How do you mean

[digital rock]

sin(tan^-1(1.1)) and cos(tan^-1(1.1)) can be simplified

[steep fox]

That is what I was asking

How so?

[digital rock]

turns out to actually be equivalent to the pythagoras method

[steep fox]

Sorry if I wasn’t clear before

Elaborate

[vital dome]

lol

[digital rock]

you want the sin of the angle whose tan is 1.1, correct?

[steep fox]

Yes

[digital rock]

http://puu.sh/Axnme/40c2d6e545.png

tan(x) = 1.1, pythagoras will tell you that the hypotenuse is sqrt(221/100), so sin(tan^-1(1.1)) = 11sqrt(221)/221

[steep fox]

Which is

It’s 11

[digital rock]

messed something up in my head :^)

[steep fox]

Nevertheless

You multiply that by 34 to get the size of one side

[digital rock]

it's about 0.7

[steep fox]

Which is seen in my first equation

[digital rock]

sure

but my point is that you have to do this step to get a simplified answer anyway

[steep fox]

You lose either way mudkip, there is no shorter way!

You can just plug my answer into a calculator?

[digital rock]

doesn't get you the exact value

[steep fox]

What

[digital rock]

well you tell me what answer the calculator gives you

[steep fox]

575.384615

Which is the correct answer

No?

[digital rock]

it's close, but it's not the exact answer

[steep fox]

What’s the exact answer

577?

Because that is most certainly NOT the answer !

[digital rock]

127160/221

oh uh, fair enough

turns out that is exactly the same, fair enough

[thick raptor]

==sqrt(2.21)

[granite stirrup]

sqrt(221)/10 = 1.48660687473185

[digital rock]

but generally it doesn't come out rational :^)

[thick raptor]

==11/sqrt(221)

[granite stirrup]

11×sqrt(221)/221 = 0.739940073395944

[steep fox]

Therefor mudkip

You admit defeat

[thick raptor]

^ that's the answer you get by finding the hypotenuse

:l

[steep fox]

Yea, my calculator agrees

[digital rock]

lmao

your expression is not the answer tho :^)

[steep fox]

What

Do we have to do this again

[digital rock]

if something asks you to find the exact value

[thick raptor]

Hm?

[steep fox]

How in heaven’s holy name is (34 x sin(tan^-1(1.1))) x (34 x cos(tan^-1(1.1))) not the answer

[thick raptor]

MathBot outputs symbolic answers now Mudkip

[digital rock]

you have to find the exact value :^)

[steep fox]

So what is the exact value

That is the exact value

[thick raptor]

🤦

feels ignored

[steep fox]

How is 575.38xm^2 not the area

[digital rock]

7480/13

[steep fox]

cm*

[thick raptor]

@steep fox that's not the exact answer

Which is what was asked

[steep fox]

No

[thick raptor]

[digital rock]

(34 x sin(tan^-1(1.1))) x (34 x cos(tan^-1(1.1))) is exact but not simplified

[steep fox]

It literally says ‘find the area’

[digital rock]

"the area" is exact but also not simplified

[steep fox]

I wrote it I should know what was asked

Okay

So how would I arrive at the exact value

[digital rock]

lmao I've literally explained the entire process twice

[steep fox]

Back to pythagoras

[thick raptor]

oof

[steep fox]

Ok so

[digital rock]

if you want to say "I plug it in my calculator", sure, but I'm not an engineer

[steep fox]

What you’re saying is that

One side is 1.1x bigger than the other

[thick raptor]

Also do hate approximate answers cuz it leads kids to rely on their calculators

[steep fox]

Which in my eyes is a coincidence as that was not what I intended

However I divulge

So by working out x you also work out y

[thick raptor]

unless the approximations are analytic/with error

[steep fox]

Which is 1.1x

So you work out x by

[digital rock]

substituting into pythagoras and solving

[steep fox]

(x)^2 + (1.1x)^2 = 34^2

[digital rock]

yeah

[steep fox]

Ok so

34^2 = 1156

[digital rock]

yeah

1.1 = 11/10 makes the LHS more doable

[steep fox]

So 2.1x^2 = 1156

Aka 21/10x^2 = 1156

[digital rock]

1 + 1.1^2x^2 = 1156

[steep fox]

Aka

[digital rock]

2.21x^2 = 1156

[steep fox]

This is a bit hard to write out in text

But

[thick raptor]

Where'd the 2.1 come from?

[steep fox]

x = 34sqrt210/21

[thick raptor]

But where did the 21 come from?

[steep fox]

1x + 1.1x

??

[digital rock]

it's x^2 + (1.1x)^2 rather than x^2 + 1.1x^2

[thick raptor]

^

[steep fox]

That is

Anyways

Final answer

After plugged into calculator

Is

23.46

[thick raptor]

Still confused

How'd you go from x^2 + (1.1x)^2 to 2.1x^2?

[digital rock]

by not squaring the 1.1

[thick raptor]

[vital dome]

[digital rock]

x^2 + (1.1x)^2 = 34^2

x^2 + 121/100 x^2 = 34^2

221/100 x^2 = 1156

x^2 = 115600/221

[steep fox]

Ok so

You have

Lemme write it down

[digital rock]

y^2 = (1.1x)^2 = 121/100 x^2 = 1156*121/221 = 139876/221, so xy = sqrt(115600*139876/221/221) = sqrt(115600*139876)/221 = 340sqrt(139876)/221 = 20sqrt(139876)/13 = 20*374/13 = 7480/13

[steep fox]

That lotta ugly

[vital dome]

the answer is a lot nicer than yours

[digital rock]

=575.(384615) recurring

[steep fox]

Am I right mudkip

Did I make any mistake

[viscid thistle]

Hi gomezz

[vital dome]

yo

[digital rock]

lhs on second line isn't 2.1x^2

[steep fox]

?

[digital rock]

(1.1x)^2 = 1.1^2 x^2 = 1.21 x^2, right?

[steep fox]

What

[digital rock]

wdym what

[steep fox]

Oh

My bad

Yea

So why does this work

[digital rock]

why does what work?

[thick raptor]

tip: when someone asks you how you did a step, it probably means you wanna check yourself

[steep fox]

x = 1x

1.1x = 1.1x

1.1x + x is not 2.1x?

[digital rock]

1.1x + x is 2.1x

[thick raptor]

2.1x

[digital rock]

(ab)^2 = a^2 b^2

(1.1x)^2 = (1.1x)*(1.1x) = 1.1*1.1 *x*x = 1.1^2 x^2

[thick raptor]

@steep fox put it shortly, you are solving the wrong problem correctly

[steep fox]

Hmm

Okay I see

So mudkip

How do you solve x^2 + (1.1x)^2?

[digital rock]

x^2 + (1.1x)^2 = 34^2

x^2 + 1.21 x^2 = 34^2

2.21 x^2 = 34^2

[steep fox]

You lost me at second step

1.21 x x^2

?

[digital rock]

it's 1.21 times x^2, yes

because 1.1^2 = 1.21

[steep fox]

Where did 1.21 come from

Oh i seeee

but shouldn’t it be x^2 + 1.21 then

[thick raptor]

but it's (1.1x)^2

not just 1.1^2

[steep fox]

Oh

Harhah yes

[digital rock]

just to make sure

you understand what 1.1 is right?

[steep fox]

Yep

[digital rock]

ok good

and 1.21?

[steep fox]

It would be easier to read on paper

[digital rock]

ok

[steep fox]

But yes what you are saying is

[digital rock]

I'll write it out on paper for you

[steep fox]

2.21 times x^2 = 34^2

[digital rock]

yes

just give me like

5 minutes to write it out

ok, I've written it out

[steep fox]

Sweet

[digital rock]

so we have 2.21 x² = 34²

so x² = 34²/2.21

and then since we have y = 1.1x, using (ab)² = a²b², y² = 1.1²x² = 1.21x²

so y² = (34*1.1)²/2.21

[steep fox]

Yep

Okay

[digital rock]

so (xy)² = (34²*1.1/2.21)²

so xy = 34²*1.1/2.21

xy = 34²*110/221

xy = 68*110/13 (cancelled factor of 17)

xy = 7480/13

[steep fox]

where did the /100 after /13 come from