precalculus

[viscid thistle]

63.5 but correct me if I am not reading the problem right

[sick steppe]

Vieta's results say that roots one at a time = -b/a

You're not asked to find every root, you're asked for the sum of all 5

[viscid thistle]

what

[dull peak]

yea but there are no other roots

except for 1

[sick steppe]

yeah, so you use vieta's results

sum of the roots one at a time = -b/a (where 1st coefficient is a, 2nd is b, etc)

[dull peak]

so its not 1

[sick steppe]

no

cause there are 4 other roots, they're just complex

1 is the only real root

,w 4x^5 +3x^4+9x^3-8x^2+16x-24=0

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by moshill1#9771

[dull peak]

eh @sick steppe its too late i already submittted it like a while ago

thanks for ur help tho

[sick steppe]

np

[blissful gale]

can someone help

please

[viscid thistle]

@blissful gale for the domain, i think the base of a log has to be greater than 1, and the argument must be positive

for solving it, know that $log_a(b) = x$ means that $a^x = b$

[obsidian monolith]

Shen

[blissful gale]

$x^{4 \sqrt{x}}=(\sqrt[3]{x})^{\sqrt{x}}$

[obsidian monolith]

Soumil

[blissful gale]

someone help me find the domain and answer

[stuck lark]

@viscid thistle idk wym "restricted at pi/2 n" for arctan

if the domain of tan is restricted to (-pi/2,pi), it becomes invertible with R as its image. arctan is defined as its inverse, so arctan has domain R and image (-pi/2,pi/2)

[obsidian monolith]

Shen

[viscid thistle]

@blissful gale natural logs is the best way to go

[dark trout]

I’m not quite sure how to use this

Do I just send a picture of my question?

[fluid sable]

Hey

[cunning hull]

Anyone help with?

[viscid thistle]

m is a vertical shift, all it will do is move your graph up and down

this is probably a question youll have a graphing calc for so graph the function 4x^3 - 3x

if you shift the graph up, you may reduce the number of roots

[cunning hull]

Wait I did something calculations first

[cunning hull]

We can't use a graphing calculator sadly

[viscid thistle]

okay

solve the equation 4x^3 - 3x = 0

uhhh

well

hold on

i dunno you'd have to solve for the turning points of the graph somehow

which isnt precalc anymore

the graph looks like thi

[cunning hull]

I tried getting the table of variation for 4x^3-3x-m

[viscid thistle]

if m>1 then you only have 1 solution

wait sorry its -m so

[cunning hull]

Wait what happened to the -m in the original equation?

[viscid thistle]

it's a shift

vertical shift

i've graphed the base function

[cunning hull]

oh ok

[viscid thistle]

m will just move this graph up and down and you can see at which points the # of solutions will change

[cunning hull]

oh

[viscid thistle]

but if you don't have a graph calc idk how to solve it

without using calculus

[cunning hull]

The part that threw me of is the (Discuss graphically)

[viscid thistle]

yeah

[cunning hull]

I'll try drawing the function one time then copying it on different values of m

thanks for the help man

[brazen tundra]

i’m not sure how to get started on this problem and confused how to find the area height or width

<@&286206848099549185>

[brazen tundra]

hello

[viscid thistle]

oh hi

[brazen tundra]

could you help me

[viscid thistle]

I think so yeah

[brazen tundra]

okay thank you

[viscid thistle]

https://www.desmos.com/calculator/cwci8ybro9 so this is the graph of y=11 sinx, but it doesn't look like the graph in the diagram 🤔

[brazen tundra]

it’s in radians

[viscid thistle]

oh, so it would really be $y=\frac{11\pi}{180}\sin\left(\theta\right)$

[obsidian monolith]

Jitter

[brazen tundra]

yeah i think so

[viscid thistle]

In that case, we know that the curve starts at 0 and goes to $\pi$ and there is a local maxima at $\pi/2$

[obsidian monolith]

Jitter

[brazen tundra]

okay yeah then what

[viscid thistle]

we know that the rectangle is going to cut the arc twice, at an equal length from the distance

so the rectangle will go from $\pi/4$ to $3\pi/4$

[obsidian monolith]

Jitter

[brazen tundra]

yeah that makes sense

i got it now it’s alright jitter

thanks for your help i appreciate it

[ancient rampart]

is sin^2(2x) the same as sin(2x) * sin(2x)?

[viscid thistle]

Yes

[ancient rampart]

ok

[ancient rampart]

can someone help with my problem

cos^2(2x) - sin(2x) - 0.8 = 0

I changed cos^2(2x) to 1 - sin^2(2x)

now I have sin^2(2x) + sin(2x) -0.2 = 0

and im stuck

[blissful ridge]

Let sin²(2x)=t and solve the Quadratic

[ancient rampart]

huh

t + sin(2x) - 0.2 = 0?

[blissful ridge]

Wait, I made an error

Let sin(2x)=t

[ancient rampart]

oh

so t^2 + t -0.2 = 0?

[blissful ridge]

Yes

[ancient rampart]

ok and then quadratic formula and then plug back in sin2x and do the double angle identity

[blissful ridge]

If you want to find the x then don't even need to do the double identity

[ancient rampart]

why not

[blissful ridge]

I mean if sin(2x)=a
Then 2x=arcsin(a)
x=arcsin(a)/2

[ancient rampart]

oh

yea

idk if im supposed to use a calculator'

probably

i get t = 0.171 and t = -1.171

but i cant take arcsin of -1.171

[blissful ridge]

Leave that

That's invalid

[ancient rampart]

but i need another answer

i should have 2 answers

the answer i get when using t = 0.171 is correct

but idk where to get the other answer from

[blissful ridge]

You'll get infinite answers

Sine is periodic

[ancient rampart]

yes ik

but 2 answers

that both have +pi(k)

[blissful ridge]

,w t²+t-0.2=0

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by The Godfather#1908

[blissful ridge]

Show me the question

[ancient rampart]

i dont have my textbook with me

i have it written down tho

idk the specific wording

[blissful ridge]

Are they asking you between some interval??

[ancient rampart]

cos^2(2x)-sin(2x)-0.8 = 0

find all solutions

roundn to nearest thousandsth

[blissful ridge]

Find all solutions would be an infinite set of solutions

[ancient rampart]

yes but its 2 solutions each with + pi(k) added to the end

each period has 2 solutions

[blissful ridge]

Show me the solution

[ancient rampart]

1 sec

when i put it into online calculator thing it says x

0.08583112+πk
4.62655786+πk

i got 0.086 + pik using one of the solutions from quadratic equation

[blissful ridge]

Are you getting this from an online calculator??

[ancient rampart]

from some website i found

i googled the question and got that

you cant get help with tests

[novel mango]

oh woops

[ancient rampart]

👍

[novel mango]

what about homework?

[ancient rampart]

hw is fine but dont expect people to just hand the answers to you

they can help but dont ask people to do it for u

[blissful ridge]

If $sin(x)=sin(\alpha)$
$\$

Then $$x=nπ+(-1)^n \alpha$$

[obsidian monolith]

My Army is Frozen

[ancient rampart]

huh

[blissful ridge]

That's how you get all the solutions

[ancient rampart]

i dont follow

[blissful ridge]

Where are you confused??

[ancient rampart]

how does that help me figure out how to get the second solution in a period

or is there only 1

[blissful ridge]

Keep changing n

And get all the values you want in your interval

Btw, n belong natural numbers

[ancient rampart]

ok

[viscid thistle]

can anyone help?

[viscid thistle]

What have you tried or what are your thoughts so far @round quest

[round quest]

i have tried 12/8

it doesnt work

[sick steppe]

I think they meant for the question you posted

not the one you posted after

[round quest]

never mind

i go tit

i got it

[lime cargo]

@round quest chaging to polar coordinates and ez

[sick steppe]

Or... S=theta * r

[round quest]

i just simplified it

3/2

[lime cargo]

But can you do that with the angle? Dont think thats legal

[bronze star]

Formula is $area=\pir^2\frac{central angle}{2\pi}$

[obsidian monolith]

AV8312

[sick steppe]

how is that not allowed exactly... it's a fraction

[lime cargo]

Ok i dont know how to read

[viscid thistle]

How do I solve this, I completely forgot the formula

[stray grotto]

Well each minute its mass will multiple by (1-0.079)

1 hour is 60 minutes, so this multiplication will compound 60 times

Ie, (1-0.079)^60

This will give us how much of the original mass remains expressed as a decimal

Multiple by the original mass, 200 to get the answee

You understand everything? @viscid thistle

[viscid thistle]

so P(1-R)^t?

I was thinking of using the natural decay rate formula, Pe^(rt)

[dire raptor]

Can a function be differentiable at a point, if the point is an end point?

My reasoning is that it is not differentiable

because for a function to be differentiable at a point

it must be differentiable on both sides of that point

[lime cargo]

What set are we talking about? A generic one in wich that point is the frontier?

[viscid thistle]

Great question.

Is it a pathological case? First, have you seen a definition of differentiability at a point?

[cobalt swallow]

@dire raptor Yes, you are right. Function cannot be differentiable at an end point, it's because it is not continuous at the end points

[serene river]

is anyone here

[vapid mica]

cos(3x+15) = .8, [0, 360)

i get two solutions, but mathway says theres 6

anyone??

[vapid mica]

nvm

actually i still have a question if someone wants to help

[onyx saffron]

i need help with converting a polar equation (a rose) to rectangular form if somebody could help

the equation is r=4cos(4θ)

[onyx saffron]

<@&286206848099549185>

[arctic kestrel]

how do I solve this???

🥺

[bold meadow]

https://discord.gg/physics

[worn anvil]

Accidently posted in the wrong channel. But I've been stumped on this. Could I get some help here?

[sonic iris]

I need help

[small bobcat]

graph appears to be -8/(x-1) +2

as you know the asymptotes are x=1 and y=2

[livid carbon]

Can anyone help with all this

[civic bone]

pick a question, and ill help

im not doing 20 questions for you

[half orbit]

im not doing hw for someone lol

[livid carbon]

Could you do 1 from each section so I can see how its done

I just need to see the structure and then I can do the rest

[civic bone]

ill do one lol, cba to do more than that

[livid carbon]

#1

[neon mural]

ah

reducible quadratics

[civic bone]

so move all the trig terms to one side

[livid carbon]

Mhm

[civic bone]

$4 \sin^2 \theta = 1$

[obsidian monolith]

Maxim

[civic bone]

\sin \theta = \pm 1/2

[livid carbon]

Where is the 1 from?

[civic bone]

this is just the first question

add 6 sin^2 to both sides

[half orbit]

lol just go on khan academy.com ez]

[livid carbon]

Ohhh got it

[neon mural]

hmm

[civic bone]

$\sin \theta = \pm 1/2$

[obsidian monolith]

Maxim

[neon mural]

you could also just solve it as a quadratic

[obsidian monolith]

lofi hip hop radio

[civic bone]

that doesn't really change much lol

[neon mural]

and so $x^2=\sin^2 \theta$

[obsidian monolith]

lofi hip hop radio

[civic bone]

this is a simple quadratic, no need for substitution

so we have the positive case, where $\sin \theta = 1/2$

[neon mural]

ah well

[obsidian monolith]

Maxim

[neon mural]

may be useful for later questions

also it's \frac{}{}

[civic bone]

ye, lemme type this out now

ik, i cba tho

so $\theta = \sin^{-1} (1/2)$

[obsidian monolith]

Maxim

[livid carbon]

I'm lost

[civic bone]

ok, go back to the beginning, things got a bit jumbled

[livid carbon]

Ok

[civic bone]

you're fine with $4\sin^2 \theta = 1$ yeh?

[obsidian monolith]

Maxim

[livid carbon]

I'm at the 1=4sin^2 theta

Yes

[civic bone]

so

$\sin^2 \theta = 1/4$

[neon mural]

divide both sides by 4

[obsidian monolith]

Maxim

[civic bone]

we take the square root of both sides (and we have to consider both the positive and negative root)

$\sin \theta = \pm1/2$

[livid carbon]

Ok..

[obsidian monolith]

Maxim

[civic bone]

is that all making sense?

[livid carbon]

Yea im writing everything down

Ok

Continue

[civic bone]

ok, so we take inverse sine of each side

we'll just look at the positive case first

ie. $\sin \theta = 1/2$

[obsidian monolith]

Maxim

[civic bone]

so we have

$\theta = \sin^{-1}(1/2)$

[obsidian monolith]

Maxim

[livid carbon]

Ok

[civic bone]

there are multiple values that theta can be though

[livid carbon]

So there isn't a definite answer?

Well at first for the general solutions I got these

[civic bone]

there is, but you have more than one you need to write down

so the blue line is 1/2

[civic bone]

we're looking at all the times that sin theta is equal to 1/2

so theres one at $\frac{\pi}{6}$

[obsidian monolith]

Maxim

[civic bone]

and one at $\pi - \frac{\pi}{6}$

[obsidian monolith]

Maxim

[livid carbon]

Oh

[civic bone]

which is $\frac{5\pi}{6}$

[obsidian monolith]

Maxim

[civic bone]

we also need to do the negative case

[livid carbon]

It says 0< or equal to theta < 2pi

[civic bone]

yep

[livid carbon]

Ok thats what I put for the general

Ok great

[civic bone]

but as i said, we need to do the case when $\sin \theta = -1/2$

[obsidian monolith]

Maxim

[livid carbon]

So those in the middle would be right?

[civic bone]

no, because its not for all n

[livid carbon]

Oh I think my teacher only cares abt one case

[civic bone]

we're only interested in values in the range 0<=theta<=2 pi

as it said in the question

[livid carbon]

Yea

[civic bone]

so for $\sin \theta = -1/2$, we have $\theta = \sin^{-1}(-1/2)$

[obsidian monolith]

Maxim

[civic bone]

which we can see here

the two solutions are then $-\pi/6 + 2 \pi$, and $(\pi - - \pi/6)$

[obsidian monolith]

Maxim

[civic bone]

or in other words, $7\pi/6$ and $11\pi/6$

[obsidian monolith]

Maxim

[livid carbon]

Alright

Thanks

Let me try the next one and see if I have the hang of it

[civic bone]

nw

there are lots of examples of this type of thing on khan academy if you get stuck

[livid carbon]

Ok for the next one i ran into this

Is this a no solution?

[civic bone]

nah, you just did it wrong

-cos^2 = -1

so cos^2 = 1

[livid carbon]

shit

Ok thanks

Ok i think this is undefined

[civic bone]

no?

[neon mural]

close

[civic bone]

look at the graph of cos x

[neon mural]

cos theta = +-1

[livid carbon]

Ummmm

At 1, x is 0

I dont get it

[civic bone]

when theta is 0, what's the value of cos?

what about when theta is 2 pi

[livid carbon]

Pi/2??

[civic bone]

plug it into a calculator

[livid carbon]

Oh

1

But Arccos 1 is 0

Man this is hurting my brain

[civic bone]

no

arcos pi/2 is 0

[uncut mulch]

arcos pi/2 is 0
arccos(pi/2) isn't defined in the reals

[civic bone]

oh sorry, brains tired lol

yeah, cos pi/2 = 0, arccos 1 = 0

[severe palm]

your base function of this one is the reciprocal function f(x)=1/x. But the graph is shifted 1 unit to the right and 3 units up.

[serene river]

can anyone help

[serene river]

@here

[woven verge]

number 32

how do i do ittt 😭

[lean tusk]

anyone know how to find an exact solution for this?

[viscid thistle]

yes you just need to satisfy 2 conditions

$tan^2x = -1$ and $sin^2x = 1$ or vice-versa

[obsidian monolith]

Jitter

[queen flare]

Precalc is ez

[viscid thistle]

yes, for some it is!!!

[queen flare]

Not for all

But to me it is I guess I shouldn't have been so bold my bad

[viscid thistle]

it's fine I know what you meant

[queen flare]

Calculus 2 I heard is a pain

[viscid thistle]

r u taking pre calc rn?

[queen flare]

No

Is there a rule I must speak in the Proper channels?

[lean tusk]

well this is the precalc channel

[queen flare]

Ok so I assume yes

[viscid thistle]

misererenobis you should help spiralio with their problem if they haven't fixed it yet

[lean tusk]

yeah... I'm still not grasping it

[sick steppe]

Basically dont have a casual discussion in a help channel

[viscid thistle]

@lean tusk I figured out the answer

[lean tusk]

could you explain how you got there

[viscid thistle]

yes but I do need to mention that there is technically infinite solutions

[lean tusk]

the question specifies in [0,2pi)

[viscid thistle]

I used the unit circle to remember that $tan^2(180) + sin^2(180) = 0$ because $tan^2(180) = 0$ and $sin^2(180) = 0$, therefore their sum must also be zero

[obsidian monolith]

Jitter

[lean tusk]

ah

[viscid thistle]

you know this because $sin = opposite / hypotenuse$ and $tan = opposite / adjacent$, so when the opposite side is zero, it will determine the result for both $sin$ and $tan$

[obsidian monolith]

Jitter

[lean tusk]

tysm!

[viscid thistle]

np and gl!

[lean tusk]

wouldn't the answer here be -pi/4 or 7pi/4? I answered pi/4 thinking that there was a typo, but I could be wrong

[lucid hinge]

$\tan(x) period is \pi . \arctan(-1)=-\frac{\pi}{4}=-\frac{pi}{4}+\pi=\frac{3\pi}{4}$

[lean tusk]

shouldn't it be +2pi @lucid hinge ? since the unit circle is 0 - 2pi

[lucid hinge]

Tan is different

[lean tusk]

oh I see

[viscid thistle]

Is this right?

[jaunty bone]

Yes

[viscid thistle]

factor and simplify

[hybrid gate]

how wud i go about doing this?

[shrewd pelican]

hey guys, how do i find the horizontal asymptote of "r(x) = 2 + x+1/x-3"
so I know its 3, cuz I graphed it. But im supposed to know with just the equation and no graphing calc.

[fiery saffron]

do you know limits

[shrewd pelican]

uhh no...

[fiery saffron]

do you know polynomial division

[shrewd pelican]

i think so

yea

with the area models?

[fiery saffron]

your HA will be 2 plus the quotient, just ignore the remainder

[shrewd pelican]

ohh i see, so x/x = 1. and then its 2 + 1?

[fiery saffron]

yeah

[shrewd pelican]

so HA=3

ohhh i see

thanks a ton man 🙂

[fiery saffron]

it makes a lot more sense with limits but youll get there

[shrewd pelican]

haha yea

[livid carbon]

I need help on #8 and #12 on this page

Anyone?

[blissful ridge]

What have you tried?

[livid carbon]

Wym

These are the last problems I have left on the page

I just have no idea how to do them

[blissful ridge]

How did you the rest of them?

[livid carbon]

Well I didnt really do those

Which is why I'm trying to these hard ones so I can figure out the other ones that look easier

[blissful ridge]

Are you familiar with quadratic equations??

[livid carbon]

Familiar is kind of an overstatement

[blissful ridge]

Well, you are going to need them

[livid carbon]

Ok

[blissful ridge]

In question no. 8 let sin(theta)=t

You'll get -t+1-2t²=0

A quadratic in t

Solve for t

[livid carbon]

T?

[blissful ridge]

Then plug t=sin(theta) again

[small osprey]

t is a dummy variable

[livid carbon]

My teacher never said anything about that

[small osprey]

I mean you could solve stuff many different ways

[livid carbon]

Ummm

These are the instructions

[fiery saffron]

Is it Identities? Or solve for x

[livid carbon]

Solve for x i think

I'm trying to solve for theta and then give the general solutions (i.e. pi/3 + pi n)

Man fuc this im goin to bed

I'll finish it in the morning

[fiery saffron]

13 try getting sec on one side tan on the other and squaring since those functions square are famously related

[viscid thistle]

13 - i'd try rewriting everything in terms of sin and cos

and clear the fractions

[worn violet]

can someone help me with Find cos x if sin x = 9/10 ?

[viscid thistle]

"quiz"

[uncut mulch]

insufficient information to determine a unique value for cos(x)

[worn violet]

@uncut mulch ok

[nova swift]

@viscid thistle what grade u in?

[vapid mica]

is that precalc?

naw i dont think so lmao

idk wot class r u in

[uncut mulch]

complex numbers should be alg2

[viscid thistle]

Is this right

[uncut mulch]

no

you didn't round the first one properly
and i have nfi what you're doing for c)

[wheat plaza]

lmao

[uncut mulch]

do you know what sec is?

[uncut mulch]

wdym by turns into -72

[viscid thistle]

Because it’s in the 2nd quadrant

and cosines in the 2nd quadrant become negative

[uncut mulch]

when using this method you don't apply the sign to the argument

but to the final result

[viscid thistle]

alright

so did I do it right or do I have to redo it

[uncut mulch]

at some point you disregarded the presence of your function(s) completely

[viscid thistle]

what do you mean

[uncut mulch]

you wrote -1/72
that made zero sense

$\sec\br{\frac{3\pi}{5}} = \sec(108\deg) = \frac{1}{\cos(108\deg)} = -\frac{1}{\cos(72\deg)}$ \
since you're going to use a calculator anyway, the last step and use of reference angles isn't even needed.

[obsidian monolith]

ℝamonov

[viscid thistle]

So I had to apply the negative to the whole function

Wait so cos 108 is all I needed?

[uncut mulch]

1/cos(108°)

[viscid thistle]

And how did I round wrong

For the 1st one

[uncut mulch]

what's the result from your calc?

[quick mirage]

Ramanov the ℝealest of them all

[viscid thistle]

.9977

[uncut mulch]

is your calculator limited to displaying 4 digits?

[viscid thistle]

No but I only need 4 digits

[uncut mulch]

round to 4dp,

[viscid thistle]

.99779

[uncut mulch]

how your round depends on the value of the digit in the 5th decimal place

[viscid thistle]

.9978

alr

[uncut mulch]

its still in rad mode

[viscid thistle]

It’s supposed to be in degree?

[uncut mulch]

for a)? yes

[viscid thistle]

.97629

.9763

alr

[uncut mulch]

there's clearly a degree symbol: ° there

also please write the 0 before the decimal point

[viscid thistle]

If it’s just sin 3

[uncut mulch]

also for c) if you have access to a calculator capable of radians (which are pretty much all scientific calcs) conversion to deg isn't even needed

[viscid thistle]

Am I still calculating in degrees

[quick mirage]

@harsh wasp what have you done so far?

[uncut mulch]

no degree symbol assume its radians

unless you know the writer is being lazy af

[quick mirage]

if theres pi just assume radians^

[viscid thistle]

Well there’s no pi

[quick mirage]

unless stated explicitly otherwise

[viscid thistle]

It’s just sin 3

[uncut mulch]

like if it were special values:
30, 45, 60, 90 etc
but the ° was omitted

it would be reasonable to assume that degrees are intended

otherwise assume its radians

[quick mirage]

it could be possible that the writer might have been lazy, but looking at the context, it seems to be radians yeah

especially looking at question b

[viscid thistle]

Alright and for C could I just write sec(-1/72) or would you advise to just do the 1/108

[quick mirage]

ok...

Just plug in x=8 for f(x) to find the corresponding value

[uncut mulch]

Alright and for C could I just write sec(-1/72) or would you advise to just do the 1/108
neither of those

[viscid thistle]

You said they equal each other

[quick mirage]

whaa.. you know what secant is right?

[uncut mulch]

yes

and where the FUK is sec(-1/72) coming from?

[viscid thistle]

I converted the radian

Into degrees

108

then got reference angle

180-108

[uncut mulch]

that would lead you to
-sec(72°)

[viscid thistle]

oh yeah

for the question

[uncut mulch]

is what right?

[viscid thistle]

that would lead you to
-sec(72°)

[quick mirage]

depending on ur calculator, you might not have the sec x, but you can solve by doing 1/cos(x)

[uncut mulch]

well i did type it..

[viscid thistle]

I don’t understand

[quick mirage]

Mmateo im not gonna give u the answer lol

[uncut mulch]

do some conversions if needed to get something that you can enter into your calculator

[viscid thistle]

do I need to find the reference angle or not

[uncut mulch]

do some conversions if needed to get something that you can enter into your calculator

since you're going to use a calculator anyway, the last step and use of reference angles isn't even needed.

[quick mirage]

yeah it just asks for the value lol

to 4 dp

iirc

[viscid thistle]

alright gotcha 😭

[uncut mulch]

and since some sci-calcs have csc, sec, cot these days

you could potentially even just set the calc in rads and punch it in directly

[quick mirage]

imagine not having texas instruments 😎

[viscid thistle]

Do you think I’d receive credit for just writing 1/108 or should I write out the whole function sec(108)

[quick mirage]

where are you getting 1/108 from

[uncut mulch]

1/108
if you wrote that, you'd get a big fat 0

[viscid thistle]

k

[uncut mulch]

as implied several times

since that shows complete disregard for your functions

[quick mirage]

sec(x) = 1/cos(x)

sec(x) != 1/x

[viscid thistle]

@uncut mulch wait I think I got the degrees switched up

for a

For cos12.5

it should be .9978 rounded right

[uncut mulch]

no

[viscid thistle]

alright nvm then it’s .9763 rounded

[uncut mulch]

there's a ° symbol there,
to indicate degrees

also please write the 0 before the decimal point

[viscid thistle]

The radian and degree mode on Desmos got me mixed up

I did

Wait

Did you think that they wanted me to get the decimals for sec(180)

So I’d input it into the calc?

[uncut mulch]

sec(180)?

[viscid thistle]

I mean 108

[uncut mulch]

do some conversions if needed to get something that you can enter into your calculator

do i have to keep repeating myself

[viscid thistle]

I got -0.3090

so it becomes sec(-0.3090)

[uncut mulch]

no

what's -0.3090 supposed to be?

[viscid thistle]

108

in degree mode

[uncut mulch]

108 = -0.3090?

[viscid thistle]

yeah

[uncut mulch]

108 = -0.3090 is a false statement

[viscid thistle]

yeah but i did cos(108)

[uncut mulch]

then why the fk didn't you say that?

[viscid thistle]

i thought you knew that

[uncut mulch]

108 is a far cry from cos(108*°*)

[viscid thistle]

alright so is it right

[uncut mulch]

well cos(108°) is approximately -0.3090
that's the only thing you have going for you atm

[viscid thistle]

yeah its sec(-0.3090)

right

[uncut mulch]

no

[viscid thistle]

why

[uncut mulch]

why are you now applying the sec function to it?

what's the definition of sec?

[viscid thistle]

1/cosx

[uncut mulch]

so why are you doing:
$$\sec(108\deg) \wthonk \sec(\cos(108\deg))$$

[obsidian monolith]

ℝamonov

[viscid thistle]

1/-0.3090

so sec should be (-0.3090)

[uncut mulch]

that's NOT what

sec(-0.3090)
means

you're showing a complete disregard for functions

and all notation that comes with it

despite my status

I already did

$\sec\br{\frac{3\pi}{5}} = \sec(108\deg) = \frac{1}{\cos(108\deg)} = -\frac{1}{\cos(72\deg)}$ \
since you're going to use a calculator anyway, the last step and use of reference angles isn't even needed.

[obsidian monolith]

ℝamonov

[viscid thistle]

BUT THATS NOT THE ANSWER

[uncut mulch]

do some conversions if needed to get something that you can enter into your calculator
since you're going to use a calculator anyway, the last step and use of reference angles isn't even needed.
do i have to keep repeating myself

[viscid thistle]

WHATS THE CONVERSION I NEED DECIMAL POINTS

[uncut mulch]

since most calcs i know of don't have sec,

conversions in that context refer to expressing it in terms of cosine or whatever

[viscid thistle]

sir please

my class starts soon

[uncut mulch]

which can lead you to 1/cos(108°)
which is NOT what you keep typing:

sec(-0.3090)
for some reason

to get your answer simply get your calc in degrees mode and get it to calculate:

1/cos(108°)

[viscid thistle]

i thought i could of just did cos(108) and put the 1 over it

[uncut mulch]

yes you can

108**°**

[viscid thistle]

I got -3.2360

so sec(-3.2360)

[uncut mulch]

NO WHY

WTF

[viscid thistle]

WHATTTTTTTTTTTTT

[uncut mulch]

are you bringing sec back in

for

[viscid thistle]

because

i thought thats what I needed to for the finished answer

[uncut mulch]

no?

do you agree that
$$\sec(108 \deg) = \frac{1}{\cos(108\deg)}$$
?

[viscid thistle]

YES

[obsidian monolith]

ℝamonov

[viscid thistle]

Yes I do

[uncut mulch]

and if you tell the calculator to evaluate the 1/cos(108°) on the right side

that'd also be the value of sec(108°)

[viscid thistle]

YES

[uncut mulch]

and that's it.
also you have rounding issues again

[viscid thistle]

when I calculated 1/cos(108) in degree mode

I got -3.23606

so sec(-3.2361)

[uncut mulch]

WHY

WHY
WHY

are you chucking sec on there again?

[viscid thistle]

because is that not right

[uncut mulch]

and if you tell the calculator to evaluate the 1/cos(108°) on the right side
that'd also be the value of sec(108°)

[viscid thistle]

HOW DO I WRITE THE ANSWER

[uncut mulch]

i.e. the approximate value of the 1/cos(108°) is -3.23606

[viscid thistle]

YES

[uncut mulch]

hence the approximate value of sec(108°) is -3.23606

because sec(108°) = 1/cos(108°)

[viscid thistle]

SO UR TELLING ME ALL I NEED IS SEC(108)?

[uncut mulch]

and since sec(108°) = sec(3pi/5) because of radians and shit

YES!!! FINALLY

[viscid thistle]

OH MY GODDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

[uncut mulch]

°

[viscid thistle]

THANK YOUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU

[uncut mulch]

wlel not 180

108°

[viscid thistle]

yes mb

@uncut mulch but

if they equal eachother

how come I cant write either or

[uncut mulch]

wdym either or?

[viscid thistle]

either the -3.23606

or 108

in the function

[uncut mulch]

why do you keep trying to write -3.23606 IN the function?

[viscid thistle]

well because you said they equal eachother

so why not

[uncut mulch]

what's they

[viscid thistle]

-3.23606
or 108

[uncut mulch]

i didn't say those were equal

[viscid thistle]

so dis cool

[uncut mulch]

no

[viscid thistle]

why do you do this

[uncut mulch]

sec(108°) is not rounded to 4dp

[viscid thistle]

so it would be

-3.2360

-3.2361

[uncut mulch]

yes

[vital helm]

Helpppp

[quick mirage]

(f o g)(x) = f(g(x))

[vital helm]

Right

That part I understand

But with the logs and variables I just don't get it

[quick mirage]

g(x) = 25^k
f(g(x)) = f(25^k) = log_5(25^k)

Do you know what the five at the bottom represents?

[vital helm]

Aye the base I believe?

[quick mirage]

Yep

[vital helm]

Maybe I'm not really sure

I don't understand this unit

[quick mirage]

what do you know about log bases

[vital helm]

They have roots and grow out of the ground

Hehehehehe sorry

Literally nothing other then they can convert to exponential form and it's math alpha a to get them on a Ti-84

[quick mirage]

What do you get when you have a base "b", and you solve for log_b(b)?

[vital helm]

1?

[quick mirage]

Yep, think of log as the inverse of an exponential function

in fact, it is the inverse of an exponential function

[vital helm]

Right

That doesn't really help me solve the question though

Well it probably should but it doesn't for me

[quick mirage]

Okay, lets get to it then

Well you have a base 5 logarithm

and you want to calculate log_5(25^k)

Wouldn't it be easier if were to have the number inside the logarithm as 5^something?

[vital helm]

No clue sorry

Would it?

I'm hopelessly lost and confused

[quick mirage]

Okay, lets start smaller

Consider the function f(x) = 2^x

[vital helm]

Right

[quick mirage]

Let's just make a table of values here to show what happens to x when it is added into this function

x = -2 | f(-2) = 2^(-2) = 1/(2^2) = 1/4
x = -1 | f(-1) = 2^(-1) = 1/(2^1) = 1/2
x = 0 | f(0) = 2^0 = 1
x = 1 | f(1) = 2^1 = 2
x = 2 | f(2) = 2^2 = 4
x = 3 | f(3) = 2^3 = 8

[vital helm]

Okay

[quick mirage]

Okay, so we know that logarithms are the inverse of exponential functions

Do you know how the inverse of f(x) and f(x) are related?

[vital helm]

The appear to be the same

[quick mirage]

Is there any relation between the base of the log function and the base of the exponential function?

[vital helm]

Uhhh is there?

I don't know

[quick mirage]

They would indeed have to be the same

Furthermore, you know that an inverse function is able to undo another function, in this case, the logarithm is able to undo the exponential function.

[vital helm]

You can do that?

[quick mirage]

That's what inverse functions do!

[vital helm]

Huh

Interesting

[quick mirage]

if g(x) = log_2(x) and f(x) = 2^x then

g(f(x)) = x

AND

f(g(x)) = x

[vital helm]

Right okay

Not sure I understand how this applies to the question

[quick mirage]

Here's what I mean

If we were to have 5^n, where n was some number, we can solve the base 5 logarithm easily, because they are inverse functions. You would get n back.

Specifically, log_5(5^n) = n

This apples to all logarithms except log_1

and negative / zero base logarithms

[vital helm]

Right right

[quick mirage]

Now, can we make 25 into some form of 5^x?

[vital helm]

So would that mean the answer to my question would be 25^k?

I mean yes you could make 25 = 5^2

[quick mirage]

Exactly, now you have:
log_5((5^2)^k)

You may have learned that (a^b)^c = a^bc

[vital helm]

Uh no I did not

[quick mirage]

This is an exponent / power rule

[vital helm]

I probably did 2 years ago but forgot

I really just am completely lost and overwhelmed by this math this semester

[quick mirage]

(2^2)^3 = 2^(2*3) = 2^6

I understand what you mean, it is stressful

[vital helm]

That's the understatement of the year

I got the test tomorrow

And more importantly if I don't get an 80 I no longer qualify for admission to the university program I selected

[quick mirage]

Try your best, and make sure that you practice when you can

[vital helm]

So essentially I'm completely fecked

[quick mirage]

Make sure you practice regularly; you gotta work hard.

Anyways, (5^2)^k = 5^2k

[vital helm]

Right

[quick mirage]

log_5(5^2k)

Since these are inverse functions, what do you get from this calculation?

[vital helm]

I have absolutely no clue

[quick mirage]

Here's a refresher if you need 🙂

[vital helm]

I'm sorry I'm just not understanding

[quick mirage]

No problem, take your time

[vital helm]

It makes no senseeeee

Hisssssssss

[quick mirage]

Ok, let me try to say less this time

lets say "b" is your log base

[vital helm]

Right right

So this question would probably take you like 30 seconds to solve but it's taken 15 minutes and I'm not even close

[quick mirage]

do not stress man, im happy to try to help

[vital helm]

I feel bad I mean your trying your best but it's me who just doesn't get it

[quick mirage]

Okay, lets make the log into a variable, so let a represent log_b(x)

b is your base

a = log_b (x)

[vital helm]

Right right

[quick mirage]

okay

now, and trust me on this

take your base

and raise it to the power of your a value

b^a

and you will get your x value

[quick mirage]

from here

for example

b = 3

x = 9

a = log_3(9)
a = 2

[vital helm]

Right so it's k^2?

[quick mirage]

It would be 2k because of power rules

(2^1)^2 = 2^(1*2) = 2^2 = 4

If you plugged (2^1)^2 into your calculator you will get the same result

(a^b)^c = a^(bc)

[vital helm]

Feckkkkkk

Inshallah I find more success tomorrow

40 minutes on one question is not good in a 25 question exam oh boy

With a time limit of an hour

[quick mirage]

Hopefully you do, I hope I was of some help. Make sure to practice a lot and ask for help from teachers, friends, or parents.

[vital helm]

Only Allah can save me

[viscid thistle]

can someone please help

idk the procedure to solving such questions

<@&286206848099549185>

[proud raven]

@viscid thistle you know what continuous means?

thats the assumption here

theyre telling you these functions are continuous, so there must be some value of a that makes it true

[viscid thistle]

so basically we need to find the value where the two functions join together

[proud raven]

yea

[viscid thistle]

yea okay

[proud raven]

then the assumption is true

[viscid thistle]

so how do we do that

[proud raven]

look at where they switch according to the PW definition

thats where they need to 'join together'

[wispy aurora]

limit x->1 2e^ax = 1?

[proud raven]

you could use a limit, im gonna guess using direct equality works too

f(a)=g(a)

something like that

[viscid thistle]

yea we didn't do calc yet

[proud raven]

solve for a

well thats a confusing choice of variables

3x^4-4 = f(x); and 2ln(ax) = g(x)

PW says they 'switch' at 2,

so solve for a by setting f(2)=g(2)

you could use desmos too

[wispy aurora]

Or you know, manipulate it algebraically

[proud raven]

you have options 😄

[viscid thistle]

how so

ughhh

[viscid thistle]

wait why x^4?

[proud raven]

bc i cant read

[wispy aurora]

mistype I'm guessing

XD

[proud raven]

anyways its literally like

check f(x) at 2

you get a value

manipulate g(x)

[wispy aurora]

Setting them equal to each other at x =2 yeah

[proud raven]

or graphically is a good option too i think

[viscid thistle]

3(2)^2 -4 = 2ln(2a)

so like this?

[proud raven]

ye

[lament elbow]

can you guys help me with this question

Find the exact values of sin((x)/(2))cos((x)/(2))and tan((x)/(2))without solving for x if secx=-8 and x is in quadrant 2. sin((x)/(2))cos((x)/(2))tan((x)/(2))

[wispy aurora]

I feel like you used waaaaaay to many brackets XD

[lament elbow]

find the exact value of sin(x/2) cos x/2 and tan x/2 without solving for x if secx =-8 and x is in quadrant 2. 1. sin (x/2) =__ 2. cos (x/2)= _ 3. tan(x/2)= _

trying to pass highschool homework ty

[wispy aurora]

Yeah this is much clearer thanks

I thought you needed to multiply them together lol

do you know about multiple and submultiple angles?

[viscid thistle]

anyone know how to do this?

[wispy aurora]

Isn't this calculus?

[viscid thistle]

this is precalc 😦

im in a pre calc class

[sick steppe]

Yeah difference quotient / AROC are pre-calc

[inner sand]

How would I go about solving this?

[uncut mulch]

remainder theorem and solve a system of equations

[ripe dust]

why the angle is below

instead of being like counterclock

hm actually i dont know if it's the good channel but uh

course : linear algebra - #linear-algebra
subject : complex stuff - #real-complex-analysis ?
according to youtube : #precalculus
@_@

here it's above

[sick steppe]

@ripe dust slightly confused on the question

[ripe dust]

oh sorry im stupid, worded it wrong

so like in the first picture the angle is taken clockwise whilst on the other one it's counterclockwise

so im just in total state of confusion

so uh is it clockwise or counterclock

[sick steppe]

So for complex numbers, the 2 main conventions for theta are (0,2pi] and [pi,-pi)

Here im assuming they are using the [pi,-pi) convention

If they went CCW then it'd be out of the allowed values, so they go CW

[wheat fox]

help pls i have a final today and dont know anything

[ripe dust]

thank you moshiii1

[serene heath]

@solid gazelle here is free

[solid gazelle]

Hoy can I develop this binomial using Newton’s thorem?

@serene heath Thank you

[viscid thistle]

is anyoneone on able to help with a question?

[orchid osprey]

do you perhaps know some trig identities like double angle

what are you allowed to use here?

[whole vessel]

Anyone know how to find the interval in which all roots of a polynomial are, without actually solving for the roots?

[orchid osprey]

hmm

yes

[whole vessel]

oh ok thanks mr bell

I'll be in contakt

[vapid mica]

can someone help?

i got 315 and 225, but photomath says its that AND 180 and 0

[odd dagger]

it's not

[blissful ridge]

Show your work

[mystic wigeon]

how do u do vector application problems with tension

this is an exampke

[cobalt swallow]

g(150 sin (62.4) + 114 sin(54.9) ) = mg

@mystic wigeon

[mystic wigeon]

how would you draw that out

[cobalt swallow]

the way it is presented exactly that way

[mystic wigeon]

what does with the horizontal mean

[cobalt swallow]

horizontal in Q mean , horizontal, same as in english

[mystic wigeon]

ok

[round quest]

how do i find the two coterminal angles for -503

[mystic wigeon]

add 360 to the angle

there are more than 2 btw

[mystic wigeon]

how do u do this problem

[vapid mica]

yo can someone check my work?

[sick steppe]

@vapid mica looks good

slight issue is that the =0 magically appears in the 3rd line

when it should be present for the whole solution

[vapid mica]

yeah sorry

but wait how are we able to solve when we have two different trig functions??

[sick steppe]

what do you mean?

[vapid mica]

i thought they had to be uniform in order to solve

so like u see the 5th line?

where i put cosx(-3 blah blah

[sick steppe]

yeah

[vapid mica]

i have a cos and csc in the same equation, but i solved anyways. i thought it had to be one trig function

like either cos or csc not both

[sick steppe]

Nope, you use the zero product property

[vapid mica]

how are we able to solve even with both??

[vapid mica]

elaborate

[sick steppe]

ab = 0 means either a = 0 or b = 0

[vapid mica]

okay makes sense

[sick steppe]

so a in this case is cosx and b is (im too lazy to type it out)

[vapid mica]

so if i plugged in pi/2, which is a solution i got from cos theta, even though i have two different trig functions, pi/2 will still work?

how does that work???

i got that solution from cosx, so how would the entire function equal zero even if i plug in for csc x?

[sick steppe]

yes, cause you get 0 times (something) = 0

[vapid mica]

oh does cscpi/2 make zero?

[sick steppe]

csc(90) = 1/sin(90) = 1

[vapid mica]

oh okay makes sense

[sick steppe]

Now let's imagine you got x = 0 as a solution from the cosx factor (it's not true but imagine)

[vapid mica]

so i can do this whenever i have: x(-3y+1) = 0

as long as i have it factored i can just set both factors equal to zero and solve?

and they dont have to be teh same trig function??

[vapid mica]

okay

[sick steppe]

x=0 would NOT be a solution of the original, since csc(0) is undefined

[vapid mica]

right

thats what im saying

[sick steppe]

so as long as all the solutions are allowed in csc, you're fine

[vapid mica]

so as long as the solutions actually work, im fine?

[sick steppe]

and yes

yeah

[vapid mica]

huh

so if i have

X(Y) = 0

i can just do

x = 0 and y -0?

y = 0

[sick steppe]

X = 0, Y = 0, and then cross check solutions with the other factor

[vapid mica]

lemme try something else

um

[sick steppe]

ie check solutions you get from Y in X and vice versa

[vapid mica]

so what if i have X+1(Y-4) = 0

y = 4, and x =-1

and then what

thats my solution?

oh wait

it is

because

[sick steppe]

(x+1)(y-4)=0?

[vapid mica]

ohhhhhhhhhhhhhhhhh

[vapid mica]

yes

[sick steppe]

yeah, x = -1, y = 4

[vapid mica]

no matter what i plug in, one will equal 0 and cancel the other one out

[sick steppe]

yes

[vapid mica]

OHHHHHHHHHHHHHHHHHHHHHHHHH

i have gone through 2 years of high school math and i have just now discovered this wow

and these rules are the same for trig functions?

[sick steppe]

Im telling you, it's the "obvious" things that people shrug off that end up being important

like 0 * anything = 0

[vapid mica]

yes

so if we cant make it uniform, just try to factor and set both equal to zero

[sick steppe]

yeah

[vapid mica]

but there will never be a time where i plug in an x, and it comes to be undefined rioght?

[sick steppe]

Like, ima say 96% of the time

you dont have to worry

[vapid mica]

because if i factor correctly, one of them will equal zero

[sick steppe]

but always a good idea to double check

[vapid mica]

one of them will have like idk a cos value of zero

[vapid mica]

double check what

factoring?

or like solutions

[sick steppe]

double check the solutions

[vapid mica]

ahhh okay i think i should be fine given this is precalc

[lunar axle]

someone plz help with any of the questions

[vapid mica]

okay makes sense thank youj

[vapid mica]

jesus mate that scared me lmao

[lunar axle]

hahahah

[mystic wigeon]

yo that was a test on schoology

wow

[gilded fern]

use identities y = r sin theta and x = r cos theta

[trail niche]

You know that ln(x) is base e

ln(b)=a implies that e^a = b

try converting the ln(b) = a functions to e^a = b functions, and see which one matches

[viscid thistle]

@plain turtle is that a test

[viscid thistle]

@viscid thistle Could you help me on a assignment?

[viscid thistle]

Just need help for part a and b

<@&286206848099549185>

[vestal zenith]

@viscid thistle which program do u use to make an exponential regression?

[viscid thistle]

@vestal zenith desmos

[vestal zenith]

uh idk about that program

[viscid thistle]

https://www.desmos.com/calculator

[vestal zenith]

I usually use TI-Nspire or wordmat

[viscid thistle]

Is not difficult