precalculus

[hard hornet]

hint: ||just add 2pi to your 2 solutions that you gave me||

[vernal spindle]

(19pi)/6 and (23pi)/6?

[hard hornet]

nice

now, remember earlier we made the substitution

$\alpha = 2\theta$

[obsidian monolith]

Cytis:

[hard hornet]

and yoru solutions are all alpha

now, solve for tehta

and there you go

[vernal spindle]

Do I have to use sin^2x+cos^2x=1 then 2sinxcosx for each one?

[hard hornet]

NO

YOu already solved for alpha

your solutions are alpha

now just divide them by 2 to get theta

[vernal spindle]

ohh

[hard hornet]

you did all the work already

now just take every solution you have, the 4 solutions, and divide them by 2

[vernal spindle]

would it be 7pi/6 * 2/1

[hard hornet]

???

literally

take the solutions you gave me

and divide them by 2

since we made the substitution alpha = 2 theta

and to solve for tehta

just divide both sides by 2

so take alph and divide it by 2

[vernal spindle]

alpha is 7pi/6 right?

[hard hornet]

alpha are the 4 solutions you gave me

[vernal spindle]

so thats one of them

[hard hornet]

ye

[warped dagger]

isnt this already simplified enough? sin^2x+cos^2x=1

says to simplify it but it seems simplified enough to me

[oblique thicket]

Write 1 = 1 or even 0=0, lol. There's no way to simplify it even further.

[narrow peak]

$sin^2x+cos^2x=1$

[obsidian monolith]

Hmm:

[grizzled totem]

If I enter this into my calculator, I get the final answer.

However, I can also do this

and then do

I'll still get the same answer

My question is, will I lose marks for not including the step of converting/simplifying eventhough I get the same answer?

[stuck lark]

ask your teacher if they want stuff simplified

[grizzled totem]

The question doesn't say to do so but the mark scheme does it anyway

I'll ask

Why is there a subtract symbol there?

[stuck lark]

they've done the laplace expansion of that determinant over the first row

[grizzled totem]

I found this during research of laplace expansions

I still don't understand why there is a minus and then a plus

[viscid thistle]

if i square sin^(1/2x) do the contents of the function sine also get squared?

to 1/4?

[willow bear]

sin^(1/2x) doesn't make any sense as stated

[viscid thistle]

sorry

i mean sin(1/2x)^2 does it become sin^2(1/4x^2)

[willow bear]

no

$\sin^2(x)$ refers to $\sin(x) \times \sin(x)$

[obsidian monolith]

Ann:

[willow bear]

it's literally just shorthand for $(\sin(x))^2$

[obsidian monolith]

Ann:

[viscid thistle]

oh okay tyvm

[stuck lark]

say you have a matrix A. we denote $A$'s entry in the $i$th row and the $j$th column as $A_{ij} $

[grizzled totem]

Furthermore, I'm watching a tutorial video on how to cross product 2 vectors. He includes add and subtraction symbols but the question I have been given just simply states numbers without add or subtraction

[obsidian monolith]

RokettoJanpu:

[stuck lark]

if $A$ is a square matrix, we define the $(i,j)$ cofactor of $A$, $C_{ij}$, as
$$C_{ij}=(-1)^{i+j}M_{ij}$$
where $M_{ij}$ is the $(i,j)$ minor of $A$, which is the determinant of the matrix created by deleting the entries in the same row and column as the $(i,j)$ entry of $A$

[obsidian monolith]

RokettoJanpu:

[stuck lark]

doing the laplace expansion of $\det(A)$ over the first row of a 3 by 3 matrix $A$, we get
$$\det(A)=C_{11}+C_{12}+C_{13}=M_{11}-M_{12}+M_{13}$$
that's where the negative sign in front of $M_{12}$ is from

[obsidian monolith]

RokettoJanpu:

[grizzled totem]

Is this my final answer?

[willow bear]

sign fuckup

should be +1j not -1j

[grizzled totem]

this?

[willow bear]

no, in the last line.

you go from -j(3-4) to -1j

-(-1) isn't -1

[willow bear]

yes that's better

[grizzled totem]

Can someone link me a video where I can learn to do this calculation

[willow bear]

look up "vector algebra" maybe? i'm sure there's plenty of videos out there

[grizzled totem]

I think it's this

except I don't know what to do with "2w" in the question I'm given, the scenario says "w" is "(-1,1,5)"

so if I have 2w, what is that? 2(-1,1,5) I'm then lost as to how to then proceed with the calculation whilst following that particulor video.

[willow bear]

2w is the vector w scaled by a factor of 2

(-2, 2, 10) in your case

[grizzled totem]

Oh

Just multiplied

[willow bear]

if $\mathbf{a} = (x, y, z)$ then $c\mathbf{a} = (cx, cy, cz)$. nothing all that complicated.

[obsidian monolith]

Ann:

[grizzled totem]

Is this my final answer?

[willow bear]

your notation is off

[grizzled totem]

Oh I did it wrong

Was supposed to be 1 - 2

Mark scheme just shows this

[willow bear]

yeah

you seem to be mixing the i,j,k notation and the ordered list notation and you're also missing out on some minus signs

which just makes everything go wack

[grizzled totem]

How would you have done it

[willow bear]

i mean i would've done what the markscheme did

like

that's how i would've written it more or less

[grizzled totem]

Is me writing the letters i,j and k irrelevent?

[willow bear]

...there is no good answer to that

[grizzled totem]

I'm watching a tutorial on how to solve this type of question

Where did he get that minus symbol from

Is it the laplace expansion thing

[willow bear]

no

AB = OB - OA

[grizzled totem]

OB is 2,1,-3 in that example

OA is -1.5.2

In this case, he's done AB = - AO + OB

Or we can just say AB = + OB - AO right?

What is that

[tawny nacelle]

lambda ?

[grizzled totem]

The markscheme shows as the letter "t"

[tawny nacelle]

oh ur writing the vector equation of a line

well thats an ugly t

[grizzled totem]

I thought it was an upside down y

[tawny nacelle]

t is used to as the parameter most of the time

[grizzled totem]

I just watched this video and it didn't even give me a final answer in an equation

My final answer is this

Mark scheme shows this

[tawny nacelle]

what r u confused about ?

[grizzled totem]

I was confused on how the equation was shown, but I get how it was done

[tawny nacelle]

those are just different ways of writing the same thing

[warped dagger]

hi folks, im in need of dire help ```

1+sin(x)/cos(x) / 1+cos(x)/sin(x) = tanx

What do i do there

not sure what to do, multiply or divide sides

[willow bear]

uh

[warped dagger]

do you have an idea

[tawny nacelle]

[willow bear]

is that meant to be this

[warped dagger]

indeed

[willow bear]

the left-hand side is a nested fraction.

do you know how to deal with nested fractions?

[warped dagger]

i multiplied numerator and denominator by sin(x)/cos(x)

[willow bear]

not a good idea. you'll still have a nested fraction after doing that.

the thing you'll want to multiply by in order to minimize the work to be done is cos(x)sin(x), so that the inner denominators are cleared

[warped dagger]

what should i do after that? do i just multiply both sides by 1 and then the 1 in the denominator cancels out or is that just wrong?

[willow bear]

do what i said first. make the fraction not a nested fraction anymore.

[grizzled totem]

What shall I search for on youtube for this type question/s?

[sleek mirage]

ohh

this is what im looking for

Complete the following sentence; make sure to cite the unit circle and what you know about sin.

The function sin is increasing in the open interval from x1 = 0 to x2 = π/2 because

what does this mean i have 3 minutes to answer it

@willow bear can you help please

[willow bear]

why are you pinging me unprompted :|

[sleek mirage]

oh my bad am i not supposed to in this server

you are the only person in this chat

who looks like they can help

**Sin is equal to Y, and on a unit circle Y is over 1. On quadrants 1 and 2, the value of Y is positive. The relationship between Sin and the unit circle is that Sin is Y, and so it is positive in quadrants 1 and 2 and negative in 3 and 4. **

[rigid beacon]

i have 3 minutes to answer

Uhhhhhh

[sleek mirage]

this is what i said

[rigid beacon]

i have 3 minutes to answer

[sleek mirage]

yes

[rigid beacon]

is this a test?

[sleek mirage]

ok maybe

[rigid beacon]

yea sorry no

[grizzled totem]

Hurry the fuck up, the exam is nearly over 🤣

[sleek mirage]

😭

[grizzled totem]

Oh sorry

[sleek mirage]

1:30 left

[willow bear]

uh yeah you attempted to cheat in an exam

[sleek mirage]

yes

is that an issue

[rigid beacon]

👋 no thanks

[willow bear]

<@&268886789983436800>

[rigid beacon]

big issue

[sleek mirage]

is that against the rules too lol

[grizzled totem]

Say you was joking or you'll be in trouble

[sleek mirage]

i was joking haha

[grizzled totem]

You're a funny guy

[sleek mirage]

its ok ill leave the server

[rigid beacon]

1. Requesting help during an exam a bannable offense.

[viscid thistle]

no joking with such things

[sleek mirage]

o

lol rip

its alright mr scientifica

[viscid thistle]

its ok ill leave the server
@sleek mirage ok let me help you with that

[sleek mirage]

ill leave the server

ok

[grizzled totem]

😦

[rigid beacon]

you realize that there are institutions that can shut down servers for this kind of thing right?

[grizzled totem]

He's gone.

[rigid beacon]

I moderate the AP server and Collegeboard has FULL power to demand that server be shut down for people using it to cheat on the AP test

[grizzled totem]

What topic is this?

Do I just search for "functions"?

[viscid thistle]

nope, I helped him quit the server by giving him a small kick

[willow bear]

yes

[grizzled totem]

?

[rigid beacon]

Ok so I've never seen that notation in a class formally but that R -> R just means it takes a real number and outputs a real number right? and that it's a 1 to 1 thing?

[viscid thistle]

yep

[willow bear]

maybe go to khanacademy

[grizzled totem]

Yes

[viscid thistle]

but doesn't mean it's a 1 to 1 function

[rigid beacon]

gotcha

[viscid thistle]

for example you can write

[rigid beacon]

but the fact that it says function means 1 to 1 right?

[viscid thistle]

f: IR -> IR

x -> x²

no we call a function 1 to 1 if for every x and y in its domain, if f(x)=f(y) then x=y

in other words, no two elements from the domain share the same image

[rigid beacon]

gotcha

but @grizzled totem as an example. Let's say h(x) = x^2 and I told you to find h(2) what would you do?

[grizzled totem]

I'd replace the x's with 2?

[rigid beacon]

yes exactly

and what would you get as your answer

[grizzled totem]

I'll go revise this and then attempt it

4

[rigid beacon]

perfect

[grizzled totem]

That's level 1 of functions

I'm going to start there

[warped dagger]

1 / sin(x)/sin(x) + cos(x)/sin(x) * cos(x)/sin(x) = tan(x)

I have no clue how to get rid off these fractions

[willow bear]

have you done what i told you to do or have you gone off in your own direction

[warped dagger]

you told me to get rid off them

i dont know how to

[willow bear]

i told you exactly what to do in clear language

i can repeat it

bolded, italicized and all-capsed

in case you didn't catch it the first time round

[warped dagger]

ok sure

[willow bear]

i want you to multiply the numerator and the denominator of the big fraction by cos(x)sin(x), and then to simplify the numerator and the denominator of the big fraction separately so that it is no longer a nested fraction.

[warped dagger]

i already multiplied it

[willow bear]

what you should end up with on the LHS after doing this is $\frac{\cos(x)\sin(x) + \sin^2(x)}{\cos(x)\sin(x) + \cos^2(x)}$

[obsidian monolith]

Ann:

[grizzled totem]

Is this right

[willow bear]

no

this is a jumble of symbols that is attempting to say something but failing to do so

[grizzled totem]

I think I know what you're saying

[willow bear]

i don't usually tend to be cryptic on that

if i wanna say something, i say it

[grizzled totem]

But it is saying that f is -2

so I replaced x with -2

[willow bear]

But it is saying that f is -2
and that already doesn't make much sense

like what are you even doing

you have the function f(x) = 4x + 2, and you're trying to calculate f(-2)?

[rigid beacon]

so 2 things I see wrong:
1: you're not finding f(x) = -2, you're finding f(-2). So that thing in the top right is wrong

2: when you sub in -2

it's 4 times x + 2

4(-2) =/= 4-2

[grizzled totem]

Is this right

[rigid beacon]

4*-2 =/= 8

4*2 = 8 tho

[grizzled totem]

oh shit

These damn mistakes

[rigid beacon]

there you go

[grizzled totem]

Is this correct

[rigid beacon]

ye

[warped dagger]

Ann:
@obsidian monolith i have a question, what happened to 1?

was that the pythogorean identity?

thats the only thing confusing me right now

still dont understand how that got there lol

i understand now, wtf

dont wanna tag you, do these just cancel out?

[odd helm]

Idk if I’m plugging my graph in wrong or I’m doing my division wrong

But could someone point out where I’m wrong

[modest flare]

The graph can be misleading. It crosses the x-axis VERY close to x=-5, but not exactly.

@odd helm

[odd helm]

Oh ok

So am I supposed to solve this problem with a calculator?

Because Idk how else to do it

[remote veldt]

@odd helm it's an application of intermediate value theorem. Check the interval boundaries. if f(x) is continuous, and f(a) and f(b) have opposite signs, f(x) has a zero in [a, b]

[thick stirrup]

Hey, I have four questions. I've spent 2 hours trying to figure out the answers to these questions but the resources provided only addresses one step of the overall equation. I really just want the Answers so I can call it a night. Could I get some help?

I need to find the missing cave angle. I know that missing portion of the big rock angle is 122 degrees.

<@&286206848099549185>

[warped dagger]

trying to solve this, havent been able to figure out how to

[serene heath]

solve in what way? @warped dagger

[grizzled totem]

Does this mean composite functions?

[wind igloo]

Yes

[grizzled totem]

I'm stuck here

I'm stuck with this question

I'm at this stage

[patent beacon]

You're done

You can simplify it, but that's the correct function

[magic abyss]

Unless they want you to operate that

[grizzled totem]

I looked at the mark scheme

They've done 3 extra steps

[magic abyss]

If that's the case then 18+2x/x+5

[patent beacon]

Looks like they did simplify it

[magic abyss]

Wait what

[grizzled totem]

Ok so I do know how they got to 8+2 (x+5)/x+5 but I have no idea how they then get 2x+18 and the one after that

Literally no idea

[patent beacon]

What's 2(x + 5)?

[grizzled totem]

2x + 10?

[patent beacon]

Yis. Then add 8

8 + 2(x + 5) = 2x + 18

[grizzled totem]

2x + 18

what about the part after that

2(x+9)?

[patent beacon]

Go the other way. See how:
2(x + 9) = 2x + 18?

[grizzled totem]

so it's saying that both of those answers are accepted?

because they mean the same thing

or is 2(x+9) the final answer

[patent beacon]

I would personally accept the one you posted originally

[grizzled totem]

Damn, I'd lose so many marks if I just did that according to this mark scheme.

[grizzled totem]

Is this correct

This is the question.

[late pewter]

no

check your algebra

the step going from the 3rd to 4th equation is wrong

[grizzled totem]

i don't know what I've done wrong

[late pewter]

$$
x = 4y + 2
$$
$$
\frac{x + 2}{2} = \frac{4y}{2}
$$

[obsidian monolith]

Frank:

[late pewter]

do you see the error

[grizzled totem]

Is that better

I inverted the plus to a minus

in stage 3

[late pewter]

yea thats right

you can always check by taking your answer and plugging it into the original equation

[grizzled totem]

My method is different to the mark scheme though

[late pewter]

i mean the methods are basically the same, you just switch the variables around at the beginning and they switch the around at the end

[grizzled totem]

How do I check my answer by taking it and plugging it into the original equation?

[late pewter]

so if you didnt switch your variables around, your answer would be
x = (y - 2)/4

right

[grizzled totem]

yes

[late pewter]

your original equation is y = 4x + 2

[grizzled totem]

correct

[late pewter]

take your answer: x = (y - 2)/4
plug it into your original equation: y = 4(y-2)/4 + 2

if you cancel everything out you get y = y

which is correct

but you always have to make sure you use your answer without the variables switched

like how they did it in the answer shown

sure but you cant do that on a test lol

ok then you do you

[grizzled totem]

I'll learn this method of checking the answer.

What would I search up on youtube to do this type of question?

[viscid thistle]

revise?

[grizzled totem]

To do this type of question

[stuck lark]

"mapping diagram"

[grizzled totem]

I had to use my eyes

Cheers

[odd helm]

To get a polynomial with zeroes 2 and 1 I would do (x-2)^2 (x-1) but I’m not sure about the -i

[uncut mulch]

conjugate root theorem

[grizzled totem]

Is this correct?

[rigid beacon]

looks good to me

you may want to label each column x and y

[maiden furnace]

I just need quick help with this small problem

https://gyazo.com/f92990fafbc1d48ec20cd07811d5cca3

how do i find the period of this

[maiden furnace]

and then does anyone have a trick to getting the 5 key points

[vocal cave]

How do I solve q1&2?

[fleet yew]

@vocal cave cant read that

Take a better pic

[fluid shore]

Okay, so what have you tried? @vocal cave

[willow bear]

uh

wow. that is

some real messy handwriting

but also, $a^b \times c^d \neq (ac)^{b+d}$

[obsidian monolith]

Ann:

[fluid shore]

@vocal cave Okay the idea here is that you want to try and get $15^x$. Notice that that's just $3^x \cdot 5^x$. So, try to get something like that on one side and bring everything else over to the other side.

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

As for the second question, the first thing you should do is to draw a picture. Once you've done that, try to get to the answer on your own and come back here if you have any problems

[harsh cipher]

Hello

Question

If the point (5,16) is on the curve y= f(x)

What point is on y= sqrt f(x-2) + 3

[fluid shore]

[willow bear]

is that $\sqrt{f(x-2) + 3}$ or $\sqrt{f(x-2)} + 3$

[obsidian monolith]

Ann:

[harsh cipher]

Hi

[fluid shore]

Lmao I feel like we ask them that sort of thing every single time

[willow bear]

use some goddamn parentheses aight

[harsh cipher]

that is

$\sqrt{f(x-2)} + 3$

[obsidian monolith]

אewb64:

[fluid shore]

Okay, so what do you actually want to find?

[harsh cipher]

I need to find what point would be $\sqrt{f(x-2)}+3$

[obsidian monolith]

אewb64:

[willow bear]

so $(5, 16)$ is a point on the graph of $y = f(x)$.

this means that $f(\underline{\phantom{33}}) = \underline{\phantom{52}}$.

fill in the blanks here

[obsidian monolith]

Ann:

[willow bear]

@harsh cipher

[harsh cipher]

yea

[willow bear]

no

i don't want a "yea", i want you to fill in the goddamn blanks

[harsh cipher]

no what

ok

f(x) = 16?

lol\

[willow bear]

no

[fluid shore]

Stop giving one word replies

Can you put a little more effort into your responses?

[harsh cipher]

Yea I can

hmm fill in the blanks

f(x) = 5?

[willow bear]

no

[harsh cipher]

what am I not understanding?

[willow bear]

you're not understanding what the graph of a function actually represents

the fact that (5,16) is a point on the graph of f means that when the input to f is 5, the output is 16

[harsh cipher]

Okay

[willow bear]

now

[obsidian monolith]

Ann:

[willow bear]

fill in the blanks here

[harsh cipher]

5 , 16?

[willow bear]

write out the actual equation with the blanks filled in

[harsh cipher]

f(5) = 16

[willow bear]

ok great

[harsh cipher]

thank you

[willow bear]

so

we have this other graph

y = sqrt(f(x-2)) + 3

and we wanna use the information we're given to find a point on this graph

notice that we're only really given the value of f(5)

[harsh cipher]

I understand that

[willow bear]

we don't know the value of f at any point other than 5

so

in $\sqrt{f(x-2)} + 3$, what value is being used as the input to $f$?

[obsidian monolith]

Ann:

[harsh cipher]

5

[willow bear]

no.

we want it to be equal to 5

but in that expression, there is no 5 anywhere to be seen

don't overthink this.

what's being plugged into f here?

[harsh cipher]

I don't know tbh

7?

[willow bear]

no

no you're overthinking this again

[harsh cipher]

haha okay

[willow bear]

you see the part of that expression that looks like f(something) right

[harsh cipher]

f(x)

[willow bear]

no

[harsh cipher]

(x-2)

[willow bear]

f(x-2)

the thing being plugged into f is, shockingly, x-2.

x-2 is being used as the input to f here.

does that make sense to you

don't overthink it

[harsh cipher]

yes

[willow bear]

ok

so

since x-2 is the input to f, and 5 is the only input for which we know the output of f

we want x-2 to equal 5

what value of x makes this happen

[harsh cipher]

x= 7

[willow bear]

ok splendid

so now

what is the value of $\sqrt{f(x-2)} + 3$ when $x = 7$?

[obsidian monolith]

Ann:

[harsh cipher]

(sqrt 5) + 3

[willow bear]

no.

what's f(5)?

it's not 5.

f(5) isn't 5.

you know what it is, but it is not 5.

you even said at some point what f(5) is

if you scroll up a bit

[harsh cipher]

f(5) out is 16

output

[willow bear]

f(5) is 16

[harsh cipher]

okay~

[willow bear]

what

are you just taking my word for it

[harsh cipher]

no

[willow bear]

so... what is the value of $\sqrt{f(x-2)} + 3$ when $x = 7$?

[obsidian monolith]

Ann:

[harsh cipher]

I need time wait please.

[willow bear]

ping me when you're back.

[harsh cipher]

okay

we said f(5) =16

so 7?

@willow bear

[willow bear]

bravo

[harsh cipher]

naisu!!

my brain is like a laggy version of 1995 ibm computer

[willow bear]

are you well-rested, well-fed and well-hydrated

[harsh cipher]

possibly not well hydrated, but I try to eat healthy

I have sleep apnea so I don't get enough oxgen to my brain as much as I would like to have

@willow bear thank you for the explanation you are awesome!

[whole hearth]

$\lim_{x\to3}\frac{\sqrt{x^2-5}-2}{x-3}$

[obsidian monolith]

mrmola:

[whole hearth]

how do I solve this?

I think once i get started i can figure out what to do but i don't know how to start

[hard hornet]

mmmm do you know the definition of a derivative?

wait first of all, do you know what a derivative is?

[willow bear]

you don't need that here

[whole hearth]

the extra fun slope thing

[hard hornet]

oh rly?

[whole hearth]

yeah i don't think i do

i am finding a limit

[hard hornet]

aight

[willow bear]

yes you can just multiply and divide by the conjugate of the num

it's a common trick

[hard hornet]

wait i'm stupid

.-. thanks Ann

[whole hearth]

i don't see how that helps though

like x+3?

[hard hornet]

conjugate of the numerator

[willow bear]

multiply and divide by sqrt(x^2-5) + 2

[whole hearth]

wait what

why?

aren't i trying to get rid of the bottom thing?

[hard hornet]

multiplying by the conjugate is a way to get rid of nasty square roots

it's a technique used for solving limits

[whole hearth]

the square root isn't the problem though

i don't understand

[hard hornet]

well, plugging in x = 3 at the moment just yields 0/0, indeterminate form, which means that we have to use some technique to get an actual answer

[whole hearth]

ooooh

i see

so this isn't a general way to solve it

its just that the problem was made in a way so that multiplying by the conjugates makes the top and bottom cancel

[hard hornet]

there are a couple ways to solve it, but Ann gave you the easiest one

[whole hearth]

but like the fact that that worked is because this is a problem designed to be possible to solve that way

what if it is a 6 instead of a 5?

[willow bear]

like under the root?

[hard hornet]

problems like these give you a chance to practice your techniques, and being able to see something, and trying it out, so when you're given a problem like the one you mentioned, you stand a chance

[willow bear]

then you'll get a nonzero number divided by 0 so the limit doesn't exist

[whole hearth]

I know to buncho

i see now

[hard hornet]

wdym

[whole hearth]

heck

like if the problem is in this form

and i do the fun conjugate thing

[hard hornet]

if it works, it works, if not, try something else

[whole hearth]

if there is no cancelling then it will be undefined?

[hard hornet]

first of all, you have to check that it's indeterminate form, then try the conjugate

if it's still indeterminate form, rip

try something else

[whole hearth]

how do i know when to stop?

when to stop trying and realize it is undefined?

[hard hornet]

this is a lot of "if" you're talking about, come back when you find a problem like this

[whole hearth]

i almost certainly won't

[vocal cave]

I don't know how to do 4aii

[whole hearth]

ty

[hard hornet]

how'd you do part ai first

[hard hornet]

part ii is almost the same, except ittells you that the two roots are numerically equal, but opposite in sign

what does that tell you the relationship between alpha and beta

[vocal cave]

What does it mean?

[hard hornet]

idk, what does it mean? you tell me

[vocal cave]

I don't know

[hard hornet]

two roots numerically equal, but opposite in sign

[vocal cave]

Does it mean the root is "3" but isn't -

[hard hornet]

it means that if one root, was, for example, 3, then the other one is -3

numerically equal, opposite in sign

if it was 6, then the other root is -6

[vocal cave]

So the root is 3 for 4aii?

[hard hornet]

if ur root was 100, then the other root is -100

no

i'm telling you what "numerically equal, but opposite in sign" means

set up the relationship between the two roots

[vocal cave]

I know that, but I don't know how to apply that logic to the question

[hard hornet]

okay, give me an equation between alpha and beta

[vocal cave]

a+b= -b/a

[hard hornet]

there's another one

[vocal cave]

ab= c/a

[hard hornet]

be careful with your variables

i'ma make it easier

[vocal cave]

o.O

[hard hornet]

let p and q be roots

p + q = -b/a
pq = c/a

[vocal cave]

Yes

[hard hornet]

there's one more relationship

between p and q

[vocal cave]

What is it?

[hard hornet]

you gotta find it

[vocal cave]

I don't know

[hard hornet]

"numerically equal, but opposite in sign"

you can find the equation between p and q with that line

[vocal cave]

-pq=-c/a?

[hard hornet]

no

you're overthinking it

if i tell you that two numbers are numerically equal, but opposite in sign, what equation can you construct that gives a relationship between these two numbers?

[vocal cave]

I don't know

-p-q=b/a

[hard hornet]

it's really simple

it doesn't involve a, b, or c

[vocal cave]

Let me think…

I've finished thinking.

I don't know.

[hard hornet]

well keep trying

you don't get answers here

you get guidance

[vocal cave]

I don't know

[hard hornet]

if i tell you two numbers, p and q, are numerically equal, but opposite in sign

what equation can you write about these two variables

it doesn't involve a, b, c

[vocal cave]

I told you what I know

[hard hornet]

keep trying, ping me back when you have something

$p + q = \frac{k-3}{4} \
pq = \frac{k+5}{4}$

[obsidian monolith]

Cytis:

[hard hornet]

there's one more equation between p and q

@vocal cave figured it out yet?

did he rage quit F

[vocal cave]

I don't know

p-q?

[hard hornet]

it's an equation

at least you're trying, i'll give it to you

p = -q

that's all

now you should be able to solve for everything

@vocal cave

[vocal cave]

I only know alpha+beta=-b/a and alpha•beta=c/a. How am I going to get alpha=-beta?

[hard hornet]

its an equation, its always true for part 2

[vocal cave]

Uh

[hard hornet]

you use that to solve for p, q and k

[vocal cave]

I still don't understand what to do

[hard hornet]

https://cdn.discordapp.com/attachments/363224154469826562/691911961088229406/20200324_152709.jpg

remember we let p and q be roots

part ii

says

the two roots are numerically equal but opposite in sign

literally means that p = -q

you NEED THIS equation to solve

part ii

[vocal cave]

I understand the question now.

But I don't know what to do

[hard hornet]

.-.

just substitute, and system of eqns and solve

[vocal cave]

Uh

[hard hornet]

you learned how to solve system of equations before?

[vocal cave]

Simultaneous equation?

[hard hornet]

yes

[harsh cipher]

okay question again

determine the equation of the radical function that has a starting point (1,3) and passes through (-3,5)

Y = ${a}sqrt{x-h}+k$

[obsidian monolith]

אewb64:
Compile Error! Click the reaction for details. (You may edit your message)

[harsh cipher]

oh no

[willow bear]

$y = a \sqrt{x-h} + k$?

[obsidian monolith]

Ann:

[harsh cipher]

naisu

yes

I need to read the manual properly 🙂

If i put in the values for (h,k) and (x,y) I get a negative number inside the sqrt.

[vocal cave]

yes
@hard hornet

Why do I get so complex roots and the answer is simply k=3?

After I get this what do I do?

[uncut mulch]

how are you getting those?

consider vietta's

[fleet yew]

Looks like symbolab

Or photomath actually

[last loom]

"Decide the primitive function F(x) to f(x) = 3x+e^x, where x=0 is 8"

I really need help with this one if a kind soul could help

[fleet yew]

@last loom a primitive function is an antiderivative

[last loom]

Yes, sorry my main language is not swedish.

is swedish*

[fleet yew]

Lol its not a term ive heard before

But you just need to integrate it

And then choose the constant of integration such that F(0)=8

[last loom]

I don't know, cant make it work. Could you perhaps help me?

Swedish is weird btw, we use some weird words 😉

Especially for math

[uncut mulch]

$\int 3x + e^x \dd{x} = \ ?$

[fleet yew]

Its e^x

[obsidian monolith]

ramonov:

[uncut mulch]

e=3

[fleet yew]

Lmao

[last loom]

Alright, makes sense

[fleet yew]

$\int 3x + e^x \dd{x} = \int 3x \dd{x} + \int e^x \dd{x}$

[obsidian monolith]

AMD:

[fleet yew]

Just practicing my latex

[willow bear]

#bots

[fleet yew]

Eh its relevant

[last loom]

The figure shows the graph of the function f, where y = f(x)

[fleet yew]

K

[last loom]

Decide with the help of the graph

Cant figure out this one either :l

[fleet yew]

What is the eq of f(x)

[last loom]

Isnt it this one?

[fleet yew]

No from the graph

[uncut mulch]

don't really need to find the equation of f(x).
what does the integral represent?

[last loom]

The integral represent the graphs area

[fleet yew]

Ye

[uncut mulch]

signed area between curve and x-axis (between x==-2 and 1)

[last loom]

So the answer should be the area in the graph that the integral says

Yeah, do you know how to calculate it?

signed area between curve and x-axis (between x==-2 and 1)
@uncut mulch

[uncut mulch]

which will be a triangle

and you can just apply the basic formula for that

[last loom]

Whats the basic formula?

[uncut mulch]

area of triagnle = 1/2 bh

[last loom]

Oh

Another one like this "Decide the antiderative to f(x) =3⋅ln(2)⋅2^3x, where F(0) =1"

It should be easy but I just cant figure it out

[willow bear]

consider that your function is $8^x \ln(8)$

[obsidian monolith]

Ann:

[last loom]

Okay

How did you calculate that?

[willow bear]

...basic exponent and log laws...

3 ln(2) = ln(2^3) = ln(8)

2^(3x) = (2^3)^x = 8^x

[grizzled totem]

Is this correct

[willow bear]

yes

[grizzled totem]

Is that how I'd write it in an exam?

[willow bear]

dunno. ask your teacher if this is acceptable on your exam to make sure

[warped dagger]

how do i get rid off these

[willow bear]

well first off sin^2(x)+cos^2(x) should ring some bells

[warped dagger]

yeah

[willow bear]

and second, nested fractions. didn't we talk about these before

[warped dagger]

alrihgt

perhaps i need to multiply both sides by cosx sinx now

[willow bear]

multiply the outer num and denom

[warped dagger]

alright

[willow bear]

also consider doing something on the right hand side

[analog relic]

I would think to expand the right hand side

And change $tanx$ to $sinx$ and $cosx$

[obsidian monolith]

ιχιση_тαкαgι:

[narrow peak]

$\frac{tanx-1}{tanx+1}$

[willow bear]

$\frac{\tan(x) - 1}{\tan(x) + 1}$

[obsidian monolith]

Ann:

Hmm:

[narrow peak]

meme

i put / wrong way round

$\frac{(\frac{sinx}{cosx}) -(1)}+1}\frac{sinx}{cosx}+1$

[analog relic]

Not (-1)

Just minus 1

But idk how to type in texit

[obsidian monolith]

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

[narrow peak]

im learning

omfg

[analog relic]

Ya i know right

[narrow peak]

fuck this shit

[analog relic]

$\frac{\frac{sinx}{cosx}-1}{\frac{sinx}{cosx}+1}$

[obsidian monolith]

ιχιση_тαкαgι:

[analog relic]

Like this?

Oh but got the ^2

So $\frac{{\frac{sinx}{cosx}-1}^{2}}{{\frac{sinx}{cosx}+1}^{2}}$

[obsidian monolith]

ιχιση_тαкαgι:

[analog relic]

Wait

So $\frac{{(\frac{sinx}{cosx}-1})^{2}}{{(\frac{sinx}{cosx}+1})^{2}}$

[obsidian monolith]

ιχιση_тαкαgι:

[analog relic]

Yay

But i rather expand with the tanx first

Then from the expansion, convert to sinx and cosx

But since we're here, let's expand

So $\frac{\frac{sin^{2}x}{cos^{2}x}-\frac{2sinx}{cosx}+1}{\frac{sin^{2}x}{cos^{2}x}+\frac{2sinx}{cosx}+1}$

[obsidian monolith]

ιχιση_тαкαgι:

[analog relic]

Then $\frac{2sinx}{cosx}=\frac{2sinxcosx}{cos^{2}x}$

[obsidian monolith]

ιχιση_тαкαgι:

[analog relic]

And $1=\frac{cos^{2}x}{cos^{2}x}$

[obsidian monolith]

ιχιση_тαкαgι:

[analog relic]

Can combine the fractions

So $=> \frac{\frac{sin^{2}x-2sinxcosx+cos^{2}x}{cos^{2}}}{\frac{sin^{2}x+2sinxcosx+cos^{2}x}{cos^{2}}}$

[obsidian monolith]

ιχιση_тαкαgι:

[pale bison]

wat in gods name

\sin and \cos btw

also \implies

[analog relic]

Since denominator is the same, just like $\frac{\frac{a}{x}}{\frac{b}{x}}=\frac{a}{b}$

[obsidian monolith]

ιχιση_тαкαgι:

[analog relic]

And $sin^{2}x+cos^{2}x=1$

[obsidian monolith]

ιχιση_тαкαgι:

[analog relic]

So => $\frac{1-2sinxcosx}{1+2sinxcosx}$

[obsidian monolith]

ιχιση_тαкαgι:

[analog relic]

Then just look at the left hand side

By multiplying $cosxsinx$ to both numbers and denominator on the left hand side, you get
$\frac{1-2sinxcosx}{1+2sinxcosx}$

[obsidian monolith]

ιχιση_тαкαgι:

[analog relic]

So therefore this is true

[grizzled totem]

I don't understand the truth table for this

The symbols mean "if mark is NOT good at maths, this means he is NOT good at computing"

Why is the third column true for this? He is good at maths, yet he is bad at computing, therefore the statement "¬r->¬s" should be false

[blazing parrot]

he is NOT good at math means he is NOT good at computing

the statement doesnt say anything about if mark is good at math

so the statement holds true

not a -> not b != a -> b

[grizzled totem]

Why is the one above it false then?

[viscid thistle]

Draw triangle labeling angles A B and C angles and a b c sides
a=3 b=4 c is unknown
Find c
can anybody help me work this problem?

[grizzled totem]

and the one right at the bottom

[viscid thistle]

i got 5 for my answer

is that right?

[blazing parrot]

2nd since r is F then by the statement s is F, but S is T so statement false

last r is T so statement doesnt apply

@grizzled totem

@viscid thistle nope

you dont know that the angle opposite side c is 90 degrees

[viscid thistle]

but do we assume it?

[blazing parrot]

not unless they tell you to

[viscid thistle]

ok

thank you very much'

😎

[viscid thistle]

if A is 25 degrees and b = 8

how do i get the third side to find the other variables?

do you know the law of sines and law of cosines?

what is it?

law of sines

now i rememeber

law of cosines

ty

actually

looking at the problem again

yeah it doesnt have another angle

I think you just need to set up a cosine or sine equation

ok

probably cosine using angle A

ok that did not format oof

so would it be cosine of 25' = 8/x

yes

ok that helps me a lot

thankiessssssssss

np

[sturdy haven]

I got -45

Please help it number two

[willow bear]

are those the only options

[sturdy haven]

Yes my frenj

[willow bear]

bc if so then the question is fucked up

none of the answer options is correct

[sturdy haven]

aight bey

is -45 right

[willow bear]

-45 is one of the infinitely many angles that are coterminal to 315

(degrees)

[sturdy haven]

you just add or subtract 360 rigjt

[viscid thistle]

to find sine i would need the hypotenuse

do i use a^2 +b^2 = c^2 or A^2 - B^2 = C^2?

[blazing parrot]

just remember in the first one c always refers to the side opposite the right angle

[viscid thistle]

kk

ty

[sturdy haven]

boi

Hello for nimber eight

Please help it

I got 3rad7/7

[blazing parrot]

for?

[viscid thistle]

your answer seems right

[sturdy haven]

my teacher is so fricken stupid bro wtf

8

She already fucked up two questions on the fucking trig review sheet

[analog relic]

7. sin240=sin(180+60)=-sin60

Third quadrant

[sturdy haven]

Yes thanks but I mean eight

[analog relic]

Oh sorry sorry

Wait ah

$sin(theta)=3/4$ so $hypothenuse=sqrt{4^{2}-3^{2}}=sqrt{7}$

[obsidian monolith]

ιχιση_тαкαgι:

[viscid thistle]

is this one B or C

[narrow peak]

pythagoras

[viscid thistle]

because i forget if the 7 goes on top

or not

[narrow peak]

then use that value and divide by 7

[analog relic]

tan(theta)=o/a=3/sqrt7=3sqrt7/7

Question 8

Oh sorry, sqrt5

[narrow peak]

,w calculate √(7²+8²)

[obsidian monolith]

Display results online and refine query

Requested by Hmm#2554

[viscid thistle]

@viscid thistle c

ok thank you

cosine should be adjacent over hypotenuse

[narrow peak]

shd be c

once u get surd at denom. u rationalise and get c

[analog relic]

$hypotenuse=\sqrt{7^{2}+8^{2}}$

[obsidian monolith]

ιχιση_тαкαgι:

[grizzled totem]

What is this in Matrices?

[analog relic]

So $cos{theta}=\frac{\sqrt{113}}{7}$

[obsidian monolith]

ιχιση_тαкαgι:

[analog relic]

Idk how to type the theta sign on texit

Do you guys know?

[viscid thistle]

$\theta$

[obsidian monolith]

Silver Hunter:

[viscid thistle]

for this we put sqr root of 13 as hypo and then put 13 over 2 and flip correct? (for cosecant)

meaning its A right?

cosecant is hypotenuse over opposite

so it should be D

oh

OOHhhhhh

ty

ur a big help

no problem!

glad to help :)

[analog relic]

Thank you @viscid thistle

[grizzled totem]

This is contradiction?

If P is not true, P is only true if q is true then it means q is true

[elfin birch]

https://prnt.sc/rlzesn

lost on what to do first here

i got it i just put it straight into my calc

[stray ibex]

convert -sin(wt+phi) to cosine , can i just put the negative inside?
so: -sin(wt+phi) = sin(wt-phi)?

[elfin birch]

ye ty ❤️

https://prnt.sc/rlzj6q

so this is just asking me to use the unit circle and write expressions that fit into that range ?

[oblique thicket]

They're probably asking you to simplify the expression.

[elfin birch]

how do i know which quadrant this is in ?

would it be Q1 ?

https://prnt.sc/rm05gu

how would i remove 2 from the theta ?

[elfin birch]

<@&286206848099549185>

[oblique thicket]

You can use double angle formula and make it $\sin(\theta)\cos(\theta) = 0.5$

[obsidian monolith]

Dmytr:

[viscid thistle]

sin(2A)=2sinAcosA

[elfin birch]

ahh ok

so whats its asking is i need to find angles on the unit circle that correspond to that angle

so i dont need to simplify

i have to get it in terms of sinx = x

so i can look on the unit circle to find angles that correspond

like pi/4 or 5pi/6

[deft fox]

pl0x help

actually i think this goes in calc

[vernal spindle]

How do I solve cosx(sinx-1)=0

I can't figure out which formula do apply next

[tawny nacelle]

well if the product of two things is 0 what can u say about those things

[vernal spindle]

one is a 0?

[tawny nacelle]

yes

either one could be 0

[vernal spindle]

What does that mean exactly?

I set one to 0?

[tawny nacelle]

both actually cuz u dont know which one is 0

[vernal spindle]

oh so find where cosx=0

[tawny nacelle]

u have to say either cos(x) = 0 or sin(x) - 1 = 0

[vernal spindle]

(3pi)/2, pi/2

since its where either cosx=0 or sinx=1?

[tawny nacelle]

hmm but if u dont have any restrictions on x, you will get infinitely many solutions

[vernal spindle]

I think we only do 0 to 2pi

[tawny nacelle]

yeah so find the x's which lie between 0 and 2pi and also satisfy one of those equations

wait maybe i should rephrase that

find the solutions for those 2 equations and make sure to only take the x's that lie between 0 to 2pi

and then those r your solutions

[gaunt bay]

hello guys, wanted to know if anyone would be able to set a time and get on a call and help me with some of my work (its a hefty amount and it would be easier than posting 50+ question every minute) all the homework is online and i also have a hard time with that too. PM would be fine and i am not expecting someone to do all the work for me. i just need a guide and someone to tell me if i got the answer right or what i did wrong. its early precalc stuff like algebra then going into trig (the class is precalc with trig) PM me if anyone would like to help.

[harsh cipher]

go through it and post questions you don't understand

Question

Equation of radical function that has starting point (1,3) and passes through the point (-3,5)

Determine the equation*

y= $a sqrt{x-h} + k$

[obsidian monolith]

אewb64:
Compile Error! Click the reaction for details. (You may edit your message)

[harsh cipher]

$y = a \sqrt{x-h} + k$?

[obsidian monolith]

אewb64:

[harsh cipher]

$y = a \sqrt{x-h} + k$

[obsidian monolith]

אewb64:

[fleet yew]

@harsh cipher your starting point always has the coordinates (h,k)

and you only have one other point

so plug the known values in and solve for a

[harsh cipher]

I got -4 inside the sqrt

I'm guessing but when the starting point passes through (-3,5) it's reflected in the y axis?

I could not solve for the value a

🤣

[uncut mulch]

was it -3 or 3,
it looked like you posted 3 yesterday

[harsh cipher]

no its -3,5

100% positivo!

[uncut mulch]

i guess in that case you'd use:

[harsh cipher]

I looked at my past notes to see if I did anything incorrect, but the questions ended up being perfect sqrt.

[uncut mulch]

$y = a \sqrt{h-x} + k$

[obsidian monolith]

ramonov:

[harsh cipher]

okay let me try that

okay I got a = 1

but under the sqrt

$sqrt-(x-1)$

[obsidian monolith]

אewb64:

[harsh cipher]

omg

$sqrt{x-1}$

[obsidian monolith]

אewb64:
Compile Error! Click the reaction for details. (You may edit your message)

[harsh cipher]

damn it

$sqrt{x-1}$

[obsidian monolith]

אewb64:
Compile Error! Click the reaction for details. (You may edit your message)

[fleet yew]

@harsh cipher

Dude its latex

It needs backslashes

Also please change ur name the right alignment is annoying

[harsh cipher]

okay

[analog relic]

$\sqrt{x-1}$

[obsidian monolith]

ιχιση_тαкαgι:

[harsh cipher]

Hi

Question

[blazing parrot]

then ask

[harsh cipher]

Determine the equations of the asymptote of the fucntion y= (3)/ (2x^2-5x-3)

[blazing parrot]

okay

so when is there an asymptote

[harsh cipher]

yes

ummm

when there are restrictions?

[blazing parrot]

um

yeah sure

so where are the restrictions

[harsh cipher]

in the denominator

[blazing parrot]

the function would have restrictions when the denominator

[harsh cipher]

cannot equal to zero

?

[blazing parrot]

the function would not be defined when the denominator is 0

so when $2x^2 - 5x -3 = 0$ stuff happens

[obsidian monolith]

Not Chezstick:

[harsh cipher]

I don't understand

[undone pawn]

do you know what an asymptote is?

[harsh cipher]

no

[undone pawn]

dont you think you should look it up before trying to solve this question..?

[harsh cipher]

yes this from the past unit I forgot how to do it. let me look it up

[viscid thistle]

An asymptote in this instance would be where the denominator is = 0.
(When you divide, it can’t equal x/0 right? So whatever values make the denominator = 0, is where your asymptotes are. Asymptotes are basically lines that aren’t strictly part of the function, but they help you graph it, and I believe that you’re trying to find vertical asymptotes right now.)