precalculus

[mellow matrix]

heeey yaa

[mellow matrix]

https://www.youtube.com/watch?v=8qMtsir0l9k

[prime depot]

i believe that is incorrect

make sure you are adding and subtracting correctly

[viscid thistle]

I double checked but i cant find an error

[prime depot]

write it out on a piece of paper the steps that you did

cuz then i can see where you went wrong

[spring thunder]

this is fking right tho

[prime depot]

is it???

[spring thunder]

yas

[prime depot]

wtf

ok hold up let me post what i did lmao cuz i wanna know where i went wrong then

[spring thunder]

mk

[prime depot]

ahhhh

lmao

k i got it

i switched numerator and denominator somehow

im gonna dissapear into the shadows now

[spring thunder]

👋

[viscid thistle]

Talk to me about this trig

are these correct?

[radiant blaze]

@viscid thistle I get cot(theta) to be sqrt(99)

[viscid thistle]

how did you solve it?

I just drew a triangle

[granite stirrup]

An internal error occurred.

A report has been automatically sent to the developer. If you wish to follow up, or seek additional assistance, you may do so at the mathbot server: https://discord.gg/JbJbRZS

[radiant blaze]

=tex $$\sin ^{2}(\theta)+\cos ^{2}(\theta) = 1(edited)

[granite stirrup]

Rendering failed. Check your code. You can edit your existing message if needed.

[radiant blaze]

sin^2(theta)+cos^2(theta)=1

so cos(theta) = sqrt(1-sin^2(theta))

and if sin(theta) = 1/10

then you can solve for cot(theta), and it's sqrt(99)

[shell salmon]

seems right

[jagged gorge]

I cant find what I did wrong

[viscid thistle]

looks about right

the method is correct, you just need to be more meticulous in your work, work it again from the start

[jagged gorge]

Ok

[mellow matrix]

wait... how did you factorise -2n^2 + 3n + 1?

you should prolly just do integration from that polynomial instead

[jagged gorge]

Oh ok

[jagged gorge]

@viscid thistle I tried it again from the start. I need to sub out 2 also?

@mellow matrix wont I need to know what a and b is tho?

[viscid thistle]

don't factor it when you integrate

that's a waste of time, integration is linear

I use the quadratic formula to get the roots,

=tex \frac{3 \pm \sqrt{33}}{4}

[viscid thistle]

scratch that I did that wrong

still, don't factor it before integrating

=tex \int_a^b -2x^2 +3x+1 dx = -\frac{2}{3}x^3 +\frac{3}{2}x^2 +x |_a^b

[jagged gorge]

So either way its still right?

[viscid thistle]

the step where you integrate it (x-a)(x-b) looks wrong

but it could be right

[jagged gorge]

Ok thx

[vocal mountain]

Ok can someone confirm the difference between something like -2^5 and (-2)^5

The former is always negative (ignore - sign and put in front after solving exponent)

and the latter can be either (odd expo. = negative, even expo. = positive) right

[distant flume]

@vocal mountain The first would have to be -(2^5) not -2^5

[digital rock]

-2^5 and -(2^5) are the same because exponentiation takes priority over multiplication (in this case by -1)

[viscid thistle]

did I solve this correctly?

[limpid plover]

=pup simplify \frac{\frac{16-x^2}{3x^4}}{\frac{20-5x}{4x}}

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@limpid plover

[limpid plover]

Yes.

[viscid thistle]

did i simplify this correctly?

[rocky bison]

If you submit your answer it'll tell you

[hexed ermine]

Yes @viscid thistle

[viscid thistle]

Can someone help me match the equation to one of the curves?

I don’t get how to do any of the questions

[winter spoke]

can anyone help me with this, Thanks

[hexed ermine]

@viscid thistle 37. -2/3x^4, it's a 4th degree polynomial

Meaning both ends will point in the same direction

Since the coefficient is negative, both ends will point down

That limits it to e f g h

[crystal current]

how? i was never taught how to do this

lol

[vale pewter]

System of equations

[crystal current]

how would each equation be set up?

[rocky bison]

@crystal current sub each value of xi into your given function

You'll only need 3 of them to solve your equation(s)

Solving for a,b and c simultaneously

[dense zealot]

@rocky bison slep

[dense fable]

hello! is this where i post a question? (:

im a little lost lol

okay, i cant screenshot my assignment right now but i'll give the word problem

the point of (-1,1) is on the graph of y=x^2. Determine the coordinates of the corresponding point on the graph of g(x)=(x+3)^2-1.

i think this is pre-calc

can someone explain this to me lol im so lost when it comes to grade 12 math.

i know since its y=x^2 its going to be a parabola, right?

so its a translation?

im thonk 2

so the ^2 is just there to show that this is a parabola, right?

yeah

so the new coordinates would be (-4,0)?

o

i got the parabola touching -4 on the x-axis???

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@zinc umbra

[dense fable]

so since it started on (-1,1), it shifts 3 units to the left so its (-4,1) then it moves 1 unit down so its (-4,0)?

idk if im doing it right lol

[hexed ermine]

Yes @dense fable

[dense fable]

i also need some help finding the domain and range of this problem. this is what my mentor sent back to me after i sent in my assignment. if that makes sense, english isn't my primary language lol

thank you! @hexed ermine

[hexed ermine]

Look at the x values

[dense fable]

and if someone can explain to me how he got y=1x1+2 from y=f(x+0)+2

[hexed ermine]

You can have any x value

So the domain is (-inf, inf)

[dense fable]

the answer is (-inf,inf)?

[hexed ermine]

The range is just all values greater than or equal to 2

Yes in interval notation

But in set builder you write it as {x€R}

[dense fable]

ooh okay, would the range be {0,4}? or is that completely off?

[hexed ermine]

No

Look at the y values

Notice that it doesnt go less than 2

[dense fable]

yes

[hexed ermine]

So the answer is {y|y>=2}

[dense fable]

ooh okay

makes sense

i do not know how i got 4 lol

also, i need help with this

Describe what happens to the graph of a function if you make these changes to its equation.

A) Replace x with x+3
B) Replace y with y-3
C) Replace x with x-2 and replace y with y+1

[viscid thistle]

A) shifts 3 to left
B) moves 3 up
C) moves 2 right and 1 down

[hexed ermine]

@viscid thistle Close

With vertical shifts, if you -1 , you go down one

[viscid thistle]

It's y sub

If U sub y-1 to y it goes up

[hexed ermine]

Oh yep, my fault, I didnt see it being replaced with y

[viscid thistle]

CHECKMATE

[hexed ermine]

Yeah I gotcha

[viscid thistle]

Get R8kt son😂🤣

🇺🇿

[sharp pagoda]

if anyone can help me with this, god bless

[prime prawn]

Ok

[sharp pagoda]

only thing i cant get my head wrapped around

[prime prawn]

problem x?

*c?

[sharp pagoda]

yeah c

i need to re sketch it with the following transformations

[prime prawn]

Whats the problem?

[sharp pagoda]

the graph of y = f(x) is shown below. sketch the graph of :

c

[prime prawn]

Alright 2 mins, I'm helping out in #prealg-and-algebra

[sharp pagoda]

alright sounds good

[prime prawn]

@sharp pagoda

Hello

[sharp pagoda]

Hey @prime prawn

okay wait so i believe i managed to answer the question

but theres another one that was bothering me more

one second

[prime prawn]

ok

[prime prawn]

Which one?

[sharp pagoda]

b

[prime prawn]

It is too blurry can you type it out please?

[sharp pagoda]

yeah for sure

[prime prawn]

14 b?

[sharp pagoda]

Describe the transformation that must be applied to the graph of f(x) to obtain the graph g(x). then, determine an equation for g(x)

so just from looking at this i can tell its a stretch and a horizontal reflection

i just need to figure out how to write this as an equation

[prime prawn]

mk

Let's look t it generally

*it

All values of f(x) are non-negative

[sharp pagoda]

yes

[prime prawn]

Whereas

All values of g(x) are not positive

what can you tell from that?

[sharp pagoda]

well its a horizontal reflextion

[prime prawn]

mhm

[sharp pagoda]

so f(?(-x?)

so -x is the first thing

[prime prawn]

no no no

the negative x doesnt go inside

it goes outside

so $$g(x) = - a f(x) + ...$$

[sharp pagoda]

oh yeah

[prime prawn]

so $$g(x) = - a f(x) + ...$$

[sharp pagoda]

-f

[prime prawn]

what else do you notice

the graph gives you a few points

For instance g(x) passes through (-7, 0) and (8, 0)

[sharp pagoda]

yes

so it stretched

its*

[prime prawn]

yes, horizontally stretched

[sharp pagoda]

okay so b is affected

[prime prawn]

b meaning?

[sharp pagoda]

-f(b(x))

[prime prawn]

yes

but -f(bx)

is that what you mean instead . of -f(b(x))?

[sharp pagoda]

yeah

[prime prawn]

Also notice that f(x) passes through points like (-1, 0) and (0,4)

these two points correspond to (-7, 0) and (3, -4) on g(x)

[sharp pagoda]

okay

so its stretched by 5?

horizontially

[prime prawn]

How did you find that out?

describe your process

[sharp pagoda]

counted

[prime prawn]

k thats one way to do it

[sharp pagoda]

is there another way you can show me

because that doesnt always work haha

[prime prawn]

Here's how to do it algebraicly

$$g(x) = a f(mx-k)+h$$

in that form

[prime prawn]

Since (0,4) corresponds to (3, -4)

you know that g(x) is shifted 3 units to the right of f(x)

if you shift units like (-7,0) 3 spaces to the left

you get (-10, 0)

with (8, 0), it becomes (5, 0)

etc.

[sharp pagoda]

left ?

[prime prawn]

Yes

[sharp pagoda]

oh nvm

[prime prawn]

Because you are shifting (3, -4) 3 spaces to the left

to get to (0, -4)

what relationship does (-2, 0) have with (-10, 0)

5 (-2) = -10

that applies to other points

like (3,-4)

[sharp pagoda]

alright

[prime prawn]

(3, -4) becomes (0,-4)

5 (0) = 0

same with the other points

do you need help with the format of g(x)

[sharp pagoda]

it will be -f(1/5(x))?

[prime prawn]

not quite

[sharp pagoda]

-f(1/5(x+3))?

[prime prawn]

test out some points

if you want to check

[sharp pagoda]

i never understood test points to be quite honest haha

okay so lets say i pick 8,0

[prime prawn]

My plan would be just to "match the points"

there are 4 points given to you in both f(x) and g(x)

you need to find the relationship between them

[sharp pagoda]

the hardest thing is seeing if it shifts left or right, thats what im most confused about

being it stretched the points move

but figuring out the shifts is another thing

[prime prawn]

ok let me see how to word this

The answer is $$g(x) = - f\left[\frac {1}{5}\left(x-3\right)\right]$$

[prime prawn]

Basically this means

that if you wanted to shift f(x) to the RIGHT

then you apply -3

[sharp pagoda]

right 3

yes

[prime prawn]

if you wanted to shift f(x) to the left 3 units

then you do +3

for horizontal shifts

its the opposite of what it seems

[sharp pagoda]

i m just confused of the whole algebriac aspect

[prime prawn]

let's take a basic function

$$f(x) = a (x-h) + k$$

[prime prawn]

what would a be

[sharp pagoda]

i forget, this was all done last year for me...

sorry

[prime prawn]

a is the dilation

k is the units moved UP

so if k is negative, you are moving down

[sharp pagoda]

oh shoot yeah

k remeber sorry dont know why that confused

[prime prawn]

h is the units moved to the right

[sharp pagoda]

but what do i do with this function, involving the points

[prime prawn]

a is -

-1

$$g(x) = -1 [b(x-h)] + k$$

[sharp pagoda]

cause of reflection okay

i call it b usually

[prime prawn]

ok

[prime prawn]

b is 1/5 because there is horizontal stretch of 5

k is 0 because there is no movement up or down

[sharp pagoda]

oh sorry i meant something else

$$g(x) = b[a(x-h)] +k$$

[prime prawn]

h is 3 because you are shifting g(x) 3 to the right

$$g(x) = b[a(x-h)] +k$$

[prime prawn]

mkay so in your format

[sharp pagoda]

okay i see

[prime prawn]

what will "b" be?

in your format

[sharp pagoda]

as you said before you took 0,4 and 3,-4 to see the shift

[prime prawn]

i was matching the dots there

[sharp pagoda]

yes

but how does this work with other dots

[prime prawn]

take (8,0)

[sharp pagoda]

okay

taken

and (1,0)

[prime prawn]

actually first let's do the main example (3,-4)

how do you get that to become (0,4)

[sharp pagoda]

like i understand you the difference of 3

but for 8,0 and 1,0 how do i figure it out

?

[prime prawn]

first do it for the basic example

how do you get from (3, -4) to (0,4)

[sharp pagoda]

-3

[prime prawn]

then?

[sharp pagoda]

well its moved left 3

from 0,4

or adding 3, which in the equation will be -3

[prime prawn]

you add 3 from the x-coordinate of f(x)

[sharp pagoda]

yeah sorry we're looking at the f(x) to g(x)

[prime prawn]

you multiply the x coordinate of f(x) by 5

then you add 3

then you multiply the y coordinate of f(x) by -1

now do that to other coordinates in f(x)

like (-2,0)

multiply 5 add 3

then multiply y coordinate by -1

[sharp pagoda]

okay i sort of understand it

[mortal rain]

y'all doing rational functions atm?

[sharp pagoda]

hey

can anyone help me quickly

transformations wise

Q: what are the coordinates of the invariant points when the function u= |x| -1

[earnest finch]

I have a logs and exponentials test tomorrow easy stuff

But then next unit is trig

disguisting

[viscid thistle]

use this one trick discovered by leonard euler to make all your trig problems into exponentials problems:

=tex \sin x = \frac{e^{ix}-e^{-ix}}{2i} \ \cos x = \frac{e^{ix}+e^{-ix}}{2}

[viscid thistle]

ezpz

[thick raptor]

lol

@viscid thistle find sin(pi/6)

[viscid thistle]

lol

draw a circle in the complex plane and cut it into equal portions of 12 ...

[viscid thistle]

Someone can help?

[prisma scroll]

You are given an arithmetic sequence. Suppose the common difference between two consecutive terms is r

Then, a_14 + r = a_15, a_15 + r = a_16

Now can you find r? @viscid thistle

[viscid thistle]

Yes I can

[sharp pagoda]

can someone explain to me how to find out if this inverse is a function or not?

equation is :

f(x) = 5/2x +3

[indigo spindle]

graph it and reflect over y=x

if it can pass the vertical line test still it is a function

[fringe stream]

an easier way would be to swap x and y, then solve for y

[crude dagger]

Why is the range of (C)(-x)(x-sqrt(2))(x+sqrt (2)) equal to (sqrt(2), 5/2) when C>0?

[blazing raven]

are you sure you copied the problem correctly?

[serene heath]

@crude dagger is there a defined domain?

if not the range is (-inft,+infnt)

[blazing raven]

(C-x)(x^2 - 2) would make more sense.

or maybe C = (-x)(x^2 - 2)

[crude dagger]

Sorry yes I forgot to mentioned the defined domain

[blazing raven]

oh. It's a quadratic.

(or a piece of one)

So if C > 0, does the parabola open upwards or downwards?

[crude dagger]

Upwards

It’s cubic right

[blazing raven]

um, btw, I think the problem has a x^2 and the solution shown has x^3 ... that might be important 😃

[crude dagger]

The x^2 on the top is another problem

[blazing raven]

fair enough.

So how many roots does the function have?

I'm guessing we can't use calculus to solve this one?

[crude dagger]

We can

[blazing raven]

oh! Are we supposed to?

That would make the problem much easier

[crude dagger]

Well no

They really don’t do anything on the solution they just say that it is as if it was trivial

[blazing raven]

er do you know derivatives?

[crude dagger]

Yes I know how to take them

[blazing raven]

okay um, do the first two lines make sense?

btw, that is a terrible usage of the word Range

They mean interval

If C > 0, then f(x) < 0 on the interval x in [sqrt(2), 5/2]

oops not square brackets

They mean interval
If C > 0, then f(x) < 0 on the interval x in (sqrt(2), 5/2)

[crude dagger]

Well the first line doesn’t hence the question lol oh okay thank you I think the rephrasing helped. I will try to work it out now. Thank you

[blazing raven]

Anyway, here's the run down. You have part of a cubic times a positive constant. The whole cubic happens to have 3 distinct roots. The restricted cubic has x between sqrt(2) and 5/2. f(sqrt(2)) = 0 for sure. (it's a root and roots are not affected by the vertical stretch C).

fair enough. You are welcome.

[limber bone]

transformations 😵

[scarlet vine]

uhh i had this question

usually math is pretty easy for me and the other questions were easy

[scarlet vine]

I dont even know if its possible; theres not even a diagram

[viscid thistle]

you need to create a diagram for yourself

oooof several hours ago

[scarlet vine]

yeah but the diagram doesnt really work out @viscid thistle

I dont think theres enough info

[viscid thistle]

i just tried it out and i agree my answer is still in terms of a variable, assuming that the electrician isn’t being forced to walk through a hallway that is only a foot wide

this is my attempt of working it out

[scarlet vine]

Apparently the answer is A

[remote musk]

@viscid thistle why'd you flip the angles in triangle 1?

shouldn't the lower angle be 30?

[viscid thistle]

the angles aren’t flipped

[remote musk]

point was shouldn't the lower angle be 30?

[viscid thistle]

of triangle 1? no.

[viscid thistle]

ooh, i’m dumb

no wonder i got stuck

[remote musk]

so assuming we can set x = anything, answer A is one of the possible lengths of the ladder, the others aren't, hence the answer is A

nah, that's a neat solution, i didn't see that

[daring pebble]

how do i find the general form for 3b?

i know it's

=tex a * \sqrt{\frac{72}{a}}^x

[daring pebble]

but how do i get to that?

<@&286206848099549185>

[rocky bison]

What is the form of an exponential function

[daring pebble]

a* b^x

[rocky bison]

ok

So now you've got a point that must lie on there

I'm assuming you know b from some other question pieces

Seeming as this seems to lead off from somewhere

[daring pebble]

??

ah yes i guess so

for f(x) = 18*2^x
b = 2
for f(x) = 8*3^x
b = 3

but how does that help find the general form?

[rocky bison]

well a and b have a relationship with one another

So if I told you that

=tex a=\frac{2}{b^2}

[rocky bison]

You want to write b in terms of a

Oof that's not correct

ok

So we have the point at x=0

=tex y=ab^{x}\x=0, y=a

[rocky bison]

Now given this we can and plug in our other point

2,72 from memory

Wait I'm still retarded

fuck my life I'm so tired

@spring thunder help

[spring thunder]

(2,72) is on the graph of that function, so

$$72=ab^2$$

[daring pebble]

ohh

[rocky bison]

and you solve for a

fuck me

[spring thunder]

^:////////

(and you solve for a or b as you wish)

[daring pebble]

thanks @rocky bison @spring thunder !

[spring thunder]

[rocky bison]

I choke so hard 😦

[scarlet vine]

wait how did mathbot get x((2sqrt(3))/3 +2) +6

@remote musk

[remote musk]

yes/

yes?*

[scarlet vine]

I keep on getting (2x(sqrt(3))+2sqrt(3))/3) + 2x

https://cdn.discordapp.com/attachments/363224154469826562/509742721913716756/latex.png

Dont know how the bot got that

[remote musk]

that's me

lol

[scarlet vine]

oh

[remote musk]

anyway

[scarlet vine]

but howd you get that?

[remote musk]

i used ririna's nice looking diagram

solved it exactly how ririna did

except, in triangle 1, you switch the 30 and 60, the lower angle is supposed to be 30

[scarlet vine]

for some reason i dont get that though

and i fixed ririnas diagram

[remote musk]

a = x/cos30 = (2sqrt(3)/3) * x
b = x+1/cos60 = 2 (x + 1)

add them up

and factor x

[scarlet vine]

oh, ive been trying to use 30,60,90 special triangle rules

[remote musk]

wait

nevermind, i edited my answers

hey did you realise I made a mistake in here?

https://cdn.discordapp.com/attachments/363224154469826562/509742721913716756/latex.png

lol

[scarlet vine]

oh

because you didnt distribute

the 3 to the 2

[remote musk]

it's supposed to be:

[scarlet vine]

yeah

[remote musk]

so when I substitute x = 3, i get answer A

[scarlet vine]

because that wouldve been 12 + 2sqrt3

[remote musk]

yeah, sorry about that, i was looking at the part where multiplied 2 by 3 to get 6, that's how 6 appeared lol

[scarlet vine]

but wait, a = x/cos30 = (2sqrt(3)/3) * x

cos30 = sqrt(3/2)

which makes it x/(sqrt(3)/2)

turning it into 2x/sqrt(3)

[remote musk]

oh nevermind it's correct

yes 2x/sqrt(3) is the same as 2 x sqrt(3)/3

[scarlet vine]

oh yeah because you rationalize then factor out x

[remote musk]

you multiply the numerator and the denominator by sqrt(3)

[scarlet vine]

thanks!

[thick raptor]

yw

[gentle flicker]

Hey guys. If i have a sine, which goes from 1 to 8 or something. How would i find the % of it that is under 4? Would i put y=4 and find the two times it crosses 4, then take whatever lengh that is call it z. then period/100 = z/%

[dim charm]

Patrick and Chris have the following conversation about their favorite numbers:
Patrick: My favorite number is an integer P > 1. It has the interesting property that 6th root(P) is an integer.
Chris: My favorite number, C, isn’t a real number! It has an interesting property, though: C^2015 = 1.
Patrick: Well, that doesn’t help me much! I can think of more than P numbers that could be the value of C.
Chris: In that case, I now know the value of C^P.
Assuming Patrick and Chris make true/perfectly rational statements, which of the following values is equal to 1

roots of unity + logic = 😦

[patent beacon]

@dim charm
What are the answers?

[dim charm]

C^3, C^5, C^13, C^31

@patent beacon

[patent beacon]

P could be 64, 729, 4096...

But there are more than P numbers C such that C is complex and C^2015 = 1. There are exactly 2014 such numbers. So either P = 64 or 729

Chris knows what C^P is from these two possibilities. As such
C^64 = C^729

Dividing by C^64 (recognizing that C ≠ 0)
1 = C^665

So, C is also a 665th root of unity.

The only factors between 665 and 2015 is 1 and 5.

But we know C itself isn't real, so C⁵ is real.

[dim charm]

thx!!

[patent beacon]

Np. Feel free to ask if you have anything else!

[viscid thistle]

I don’t get how we get the transformed points

[scarlet lance]

f(x+1) means you move the graph 1 space to left

and f(x) cross (4,0), so you put 0 at f(x+1), i guess

[gleaming bobcat]

If you are putting two intervals into a set, and they are both share a closed endpoint, are their ranges simply added, then?

I mean like a piecewise function... If two subsets, each reaching to opposite infinity, and touching at the middle (with >= and <= meeting up perfectly), are simplified, do they merge? Or do they have to share a function as well as a domain? Very confusing

to me

[limber bone]

give an

example

[gleaming bobcat]

Okay, so imagine if in this Piece-wise Function, the second (middle) function were to intersect directly with the endpoint of either function 1 or 2...

Would I use an upside down Union, an Intersection? Or is this not allowed with a real piecewise function because of vertical line test?

[limber bone]

union i think

ye union

like for example if it intersected with the end point of the first function the parabola

the parabola stops at x = 1

so for example

domain is from - infinity to 1 U from 1 to wherever the middle function ends

got it?

forgetting about the third functoin

function*

https://www.youtube.com/watch?v=0ZoSxPg6JSQ&t

here u go

[gleaming bobcat]

Ahh, Union + brackets (not parentheses), definitely cleared up my question and makes more sense, thanks!

I will watch the video later, but I just have to say Organic Chemistry Tutor saved my ass last semester in Chemistry, enough people like this guy could easily replace the Pearson-owned "education" system. Just my 2 cents.

[trim coyote]

Hey, can someone explain by b^2 = a^2 + c^2 -2accosB becomes cos(B) = (a^2 + c^2 - b^2)/(2a*c)

When I reorder, I get cosB = b^2/(a^2 + c^2 - 2ac)

[fallen cloud]

because

you subtract the a^2 and c^2 first

to bet (bb-aa-cc) = 2ac * cos(B)

then divide

[trim coyote]

ahhh....

I see. I was dividing by cosB on both sides, inversing, then moving b^2 over to the right side as the numerator

What rule did I break here?

[fading token]

The distributive law

Remember that a(b + c) = ab + ac

[viscid thistle]

How would I solve this via factoring?

[serene heath]

uh that has imaginary roots

so completing the square would be easier

[viscid thistle]

Imaginary roots? Does that mean it’s prime?

[serene heath]

no imaginary roots mean theyre in the form $$a+bi$$ where i =$$\sqrt{-1}$$

[serene heath]

if you sketch that graph

you will see that it never intersects the x axis

and hence does not have real solutions

[viscid thistle]

Ahh

[serene heath]

=pup x^2+2x+5

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@serene heath

[serene heath]

see? doesnt cut the x axis

[viscid thistle]

Ye ty

[serene heath]

are u sure its given correct?

[viscid thistle]

Yep

[prime prawn]

Can someone explain linearization on a graphing calculator to me?

I'd like to know why in a power function y = ax^b, you graph (lnx, lny)

and other examples

[lost cipher]

we can ln that equation to get:

lny = ln(ax^b)

which is:

[prime prawn]

uhh

lmo

*lmao

[lost cipher]

lny = ln(a) + bln(x)

[prime prawn]

yea

[lost cipher]

so you have y = ax+b graph

[prime prawn]

so ln (a) is a constant

ln y = lnx + ln a

hows that a supposed to be close to a line though?

[lost cipher]

you forgot b, which becomes the slope of that graph

[prime prawn]

yea mb

[lost cipher]

it is a line

[prime prawn]

b is the slope

but ln (x)

??

[lost cipher]

you treat ln(x) like x

[prime prawn]

wait nvm sorry

because you subbed in x for ln x

in the beginning

ty

[lost cipher]

np

[prime prawn]

@lost cipher

I'm confused about 2 more.

y = ax^2+bx+c, you graph (x, sqrt y)

[lost cipher]

you have an extra x thou so it doesn't become linear

[prime prawn]

?

it does

the book states

[lost cipher]

if it was like y = ax^2 it would work

[prime prawn]

"To linearize data modeled by: a quadratic function y = ax^2+bx+c, graph (x, sqrt y)"

[lost cipher]

i might be wrong, let me try an example

y = x^2 + 2x

[prime prawn]

I'm not doing this stuff; the calculator is when you give a set of data.

[lost cipher]

(-3,3),(-2,0),(-1,-1),(0,0),(1,3),(2,8),(3,15)

[prime prawn]

a bunch of points that can be best represented by by parabolas

mhm

[lost cipher]

hold on

what do you even do when y is negative?

[prime prawn]

??

the y value

idk

[lost cipher]

yeah the text is def wrong

[prime prawn]

nah

lemme model it

without the negative i believe

[lost cipher]

(0,0),(1,3),(2,8)--> (0,0),(1,sqrt3),(2,sqrt8)

the slopes between the first two points and the last two points are not the same

[prime prawn]

it wont

it'll be close though

when linearized

[lost cipher]

yeah it'll be close at the point you linearize

because you can assume bx is irrelevant compared to ax^2

[prime prawn]

yep

[lost cipher]

i thought you meant the whole thing cuz the first one you mentioned applied for the entire graph

[prime prawn]

so basically ax^2 + bx --> ax^2 as x approaches infinity

or they become closer

[lost cipher]

yeah pretty much

[prime prawn]

so if you do it with ax^2

y = ax^2

sqrt y = sqrt (ax^2)??

or y = sqrt (ax^2)

[lost cipher]

sqrt y = (sqrt a)x

[prime prawn]

yea

but

why do you apply sqrt

to both x and y

oh nvm

[lost cipher]

that's because the equation has to hold

[prime prawn]

sqrt y = (constant)x

[lost cipher]

sqrt both sides pretty much

[prime prawn]

(sqrt y, x)

ok

and for y = ab^x

it says grah (x, ln y)

so you ln both sides

nvm

[lost cipher]

yeah you're on the right track

[prime prawn]

ln y = (constant) + x (ln b)

ln b is another constant

ohh

so thats why its represented like that

thank you

[lost cipher]

np

[rare zephyr]

How to find coordinate R

[vestal widget]

eye ball it

lol

[red prawn]

http://www.wolframalpha.com/input/?i=solve+pq%3Dac,+p%2Bq%3Db+for+p,+q

this is neat

probably should have expected it to be honest though

[earnest finch]

@rare zephyr I got it but it's hard to explain

There are alternate interior congruent angles

Which make a pair of triangles similar

Then you can find the distances

And then find the slope of PQ and find the coordinate by making an equation for the point SR

Yeah too long for me to explain lol

[dim charm]

make equation for rs using slope for pq because they are parallel, and using the point (5,0). once you have that line, find the equation of the line that intersects qs perpendicular, since you are given the point and can find slope since -1/slope of qs. once you have that find where they intersect

[rare zephyr]

Thanks

[rare zephyr]

So R is the point of intersection of, let's pick QR and SR. Which means at R the value of y for both of the straight line equations are the same. So I can substitue Y for the other straight line equation. I get it now

[viscid thistle]

For the last question, the roots should be -12 +14

But my text book says +14 +12

[tawdry current]

+12 and -14 right?

[viscid thistle]

I think it should be -12 and +14

Text book thinks both negative or both positive

[tawdry current]

Huh thats weird...

I did it on paper and got +12, -14 and also my calculator says +12, -14

definitely not both positive/negative...

[viscid thistle]

Yea you’re right because after my foil meyhodnyou have to inverse to show roots....

[serene heath]

=wolf solve x^2+2x-168

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@serene heath

[serene heath]

wait just lookin at the equation

they cant both be positive

cuz last term is negative

[tawdry current]

Yes and its so large

so no way the roots would only be 2 apart.

[serene heath]

yup

[viscid thistle]

@serene heath @tawdry current ty.... guess my txtbook’s a bit autistic

[tawdry current]

Most of my books have also had those kinds of goofs around them

from my current ones I have not yet found any surprisingly

But yeah np

[earnest finch]

What does the Greek letter Sugma mean again?

[slender river]

uh

s i g m a or is this a sugma ballz joke

i cant tell

[earnest finch]

damn

reee

it was the joke lol

[slender river]

lel

[unreal plaza]

bruh what does a turning point mean
is that dy/dw = 0

[floral tide]

yeah

no thats peak point

turning point of wahat?

of the derivative from neg to pos?

[rare zephyr]

how2do

[worn prairie]

@rare zephyr remember that, for angles in radians, arc length = angle * radius

so you can set up an equation with unknown radius and known arc AB + CD

[jagged gorge]

I'm stuck how do you figure out the gradient of the line

[fringe stream]

slope of the tangent of any curve is given by its derivative

[jagged gorge]

so do i need differentiate y=x^3?

then i get the gradient?

[midnight spindle]

yes

[jagged gorge]

does derivative mean differentiate

[midnight spindle]

yes

[jagged gorge]

k thx

[midnight spindle]

no problem

[jagged gorge]

how do i find the gradient for the line?

[jagged gorge]

i got -2 but the gradient is -1

[dull imp]

@jagged gorge y=x^3-x^2-2x

dy/dx=3x^2 - 2x - 2

dy/dx | (x=1)= 3 - 2 -2

= 3 - 4

= - 1

[jagged gorge]

wait what did u do do for the second step

[dull imp]

@jagged gorge thats just the power rule

so d/dx x^n = n * x ^ (n-1)

[jagged gorge]

the one that u edited

[dull imp]

@jagged gorge oh no for there i just plugged in x=1

[jagged gorge]

y 1 ?

[dull imp]

so 3 (1)^2 - 2(1) -2(1)

= f prime (1)

or the derivative at the point x=1

so your new function is 3x^2 - 2x -2, right? and u wanna find the slope at x=1. so you just plug in 1

[jagged gorge]

so it only works for slope and not other points?

[dull imp]

@jagged gorge im not sure what you mean by that?

[jagged gorge]

@dull imp like y did u specifically sub in x=1 and not other points

[dull imp]

@jagged gorge well doesnt the question ask to find the slope at the point x=1?

judging by the image, there's a dotted line at x=1

[jagged gorge]

but thats the answer

in the original question it didnt have a graph

[dull imp]

@jagged gorge i thought u said the answer was -1?

[jagged gorge]

yes

[dull imp]

its asking for the slope at a particular point, correct?

[jagged gorge]

the question was to find the area

[dull imp]

oh ok

[jagged gorge]

but i was stuck on finding the gradient for the line

[dull imp]

could you post the original question?

oh wait

[jagged gorge]

k

[dull imp]

well if the line passes through (0,-1) and is tangent to that equation, then it doesnt necessarily have to be like that...

[jagged gorge]

ooo nvm i get it

thx

[dull imp]

uh cool lol.

[kindred violet]

hey :)

e^ ln x = ___ for each positive number x

im kinda stuck on this one :(

[serene heath]

@kindred violet thats just x

[viscid thistle]

How do you do (m)

[serene heath]

try the substitution $$\tan(u)=\sqrt{2}x$$

[viscid thistle]

But then you would have the derivative of tan^-1 (u) as the denominator

Wouldn't that over complicate things

[serene heath]

wot

oh ya

it doesnt overcomplicate it that much

[viscid thistle]

Hmm

[serene heath]

you defo will get arctan in your integral

[viscid thistle]

I c

[remote musk]

easier to substitute $$ x = \frac{1}{\sqrt2}tanu $$ isn't it?

[remote musk]

wait

nvm

it's the same

[viscid thistle]

@remote musk I just realised you can sub root 2 x =U

It gives you tan inverse function then

[remote musk]

ah, i dont have derivative of the inverse of tan memorised

[serene heath]

its 1/1+x^2

why do you need it tho

to check?

[viscid thistle]

Could someone explain to me why does this limit equal infinity?

[viscid thistle]

nvm got it u just plug a slightly bigger number than 1 and then its obvious it reaches infinity

[wild mica]

hmmthink of it as 1 + 1/(x-1)

you can check it

[viscid thistle]

how to approach?

[wild mica]

solve for (x-4) ^3/4

[viscid thistle]

how do you strech the bionomal out

[wild mica]

yo dont

what do you mean?

[viscid thistle]

wait so i do 8^4?

[wild mica]

then you take the 3th root of that

[viscid thistle]

ah k

thanks

[remote musk]

basically, raise everything to the power 4/3

[viscid thistle]

yeah

what does the notation mean

[spring thunder]

it's function composition

$$(f\circ g)(x) = f(g(x))$$

[spring thunder]

(ie put the output from g as the input in f)

[viscid thistle]

oh

thx

[spring thunder]

👌 and delete your thing from calculus

[viscid thistle]

is this correct?

Well, sin(3x) isn't the same thing as 3sinx, although the limite you take is ultimately correct

[thick raptor]

so no-ish @viscid thistle

you wanna do sin(3x)/(3x)

[viscid thistle]

damn

[thick raptor]

no

[viscid thistle]

could u show me how? im new to limits and im kinda confused how am i supposed to calculate them

SA is correct. sin(3x)/(3x) goes to 1 because sin(anything)/anything goes to one as anything goes to zero

Well 5x = 3 * 5/3 * x -- perhaps that will be helpful?

i know sinx/x equals 1 but thats it i have no idea how to get 3/5 if i cant do the thing i did on the pic

sin3x if x approaches 0 is 0

i dont know what u mean by sin(3x)/(3x)

that means sin of 3x divided by 3x itself

isnt that sin 1?

wait maybe i get it

just tell me if this is correct if not forget it

I'm doing drive-by helping, hoping someone else will also come to your aid :)

this is the original

correct or nah? 🤔

[spring thunder]

multiplying by 3x/3x is a bit too much

3/3 is already enough

$$\frac{\sin(3x)}{5x} = \frac{\sin(3x)}{5x}\cdot\frac33 = \frac{\sin(3x)}{3x}\cdot\frac35$$

[viscid thistle]

thank you

[thick raptor]

no

[spring thunder]

nome

[viscid thistle]

what is te inverse of y = (x+5)^2

[shell salmon]

to find the inverse, you are replacing the domain (x) and range (y) with each other

so in the function, you would replace x with y, y with x

after that, you would isolate y, as functions are written y=

[viscid thistle]

but why is it plus or minus sq rt

[thick raptor]

because there are two solutions to y² = stuff

(+y)² = (-y)²

[tawdry current]

is the inverse like the original function solved for x and then x and y changing place pretty much? All the operations done but in a reverse order right?

I haven't had that yet but pretty interesting.

[thick raptor]

pretty much yeah

[royal gust]

Hi can anyone help me solve this?

[native timber]

they're powers of 2

S_1 = 1
S_2 = 2
S_3 = 4
S_n = 2*S_(n - 1)```

[royal gust]

ok

so where does this s come from

[native timber]

it's just the name of the sequence

[royal gust]

its not a?

[native timber]

the first element of the sequence S is 1, the second element of the sequence S is 2 etc

[royal gust]

and also for the formula (n-1) its not squared like $$2*S^(n-1)?$$ ?

[native timber]

no, it's subscript meaning that the nth element of the sequence is equivalent to the previous element of the sequence but doubled

you could use exponents to get the nth element in this sequence using 2^(n - 1)

[royal gust]

oh ok

so what different between 2^(n-1) and what you had earily? will I get the same results?

[native timber]

yes you'll get the same results, what i had earlier just makes the sequence more "explicit" by defining elements in terms of other elements

[royal gust]

wow thats interesting its like science getting the same results in different odd ways

[native timber]

they aren't that different, exponentiation is repeated multiplication and a geometric sequence is a sequence in where an element is the previous element but multiplied by some constant

you could define the nth element of any sequence using S_n = S_1 * r^(n - 1) where S_1 is the beginning of the sequence and r is the constant that the sequence "grows" by

hold on im gonna learn how to use mathbot real quick

[royal gust]

ok

[native timber]

$$S_n = S_1 * r^{n - 1}$$

[royal gust]

wow you got it quick

[native timber]

or $$S_n = S_{n - 1} * r$$

[royal gust]

so the n was underneath

[native timber]

that's typically the notation for sequences

$$S_n$$ is the nth element of the sequence S

[native timber]

_ to denote subscripts

[royal gust]

$$S_{1} = 1$$

[royal gust]

what would be r in this case?

[native timber]

r is the common ratio and it's the constant which you multiply by to get the next element

in the sequence {1, 2, 4, 8, ...}, that r is 2 because it's doubled each time

in the sequence {10, 30, 90, 270, ...}, that r is 3 because it's tripled

[royal gust]

$$r = 2/1 = 2$$

[native timber]

yeah

$$r^x = \frac {S_n} {S_(n - x)}$$

[native timber]

$$r^1 = \frac {S_n} {S_(n - 1)}$$

[native timber]

just realized that i messed that up

$$r^x = \frac {S_n} {S_{n - x}}$$

[native timber]

much better

[royal gust]

$$r^{1} = \frac{S_2}{S(n-2)}$$

[native timber]

well no, that's not true

if we use your sequence {1, 2, 4, ...} we can assume that S_0 is equal to 0.5

$$2/0.5 /= 2$$

[native timber]

that's awful lol

[royal gust]

$$r^x = \frac {S_2} {S{2- 1}}$$

[royal gust]

how about this?

[native timber]

i guess that's true iff x is equal to 1

[royal gust]

we had 1 there earlier just trying figure out what n can be replace by

so we can solve this formula

$$s_n = 1 * 2 ^{(n - 1)}$$

[royal gust]

$$S_n = 2^{(n -2)}$$

[royal gust]

$$S_8 = = 1 * 2^{(8 - 1)}$$

[royal gust]

I think I know what n is

its the highest number isnt it?

[native timber]

what do you mean?

S_n is just the nth element in the sequence S

[royal gust]

since we have 1,2,4,8

the highest sequnce I mean

[native timber]

your question had the sequence 1, 2, 4, 8, ...

[royal gust]

right

[native timber]

... means keeps on going

[royal gust]

ohh ok

[native timber]

there is no "highest" element, it's an infinite sequence

[viscid thistle]

Well I guess there KINDA is.. infinity lul

[royal gust]

heres what I have so far

$$1 *2^7 = 2^7 = 128 = 128 S_8$$

[royal gust]

Im not sure what you mean by infinty, but we do have to find the 8th element for this one

what do we do next to solve this part? how do we solve S_8?

[native timber]

$$S_n = S_1 * r^{n - 1}$$ therefore $$S_8 = 1 * 2^7 = 128$$

[royal gust]

thats great so that was it

[native timber]

yeah

you just needed to use the equation and substitute all required values (which were S_1, r and n in this case)

[royal gust]

thats what im thinking too

I am having problems plugging which numbers in

I dont think we are done yet. So how do we solve 128_8?

[native timber]

what do you mean?

$$S \neq 128$$ but $$ S_8 = 128$$

[native timber]

S itself isn't the 8th element, it's the sequence

[royal gust]

the final answer is not 188?

[native timber]

S_8 is the 8th element (which is 128)

[royal gust]

thats the answer I got from my teacher 188

I dont know how to get it though

[native timber]

well it shouldn't be 188; it isn't even a power of 2

[royal gust]

I thought you said something about 2 eariler

[alpine portal]

Help me pls.

I also tried... (-∞,-5)U(-5,-4 1/6)U(-4 1/6,∞)

[fringe stream]

simplify f(f(x))

[solemn tiger]

Simplify the fraction you have of (f o f)(x)

Check your denominator.

[alpine portal]

x/6x+25

[fringe stream]

what will the domain of that be then?

[alpine portal]

x is still undefined at -4 1/6

[fringe stream]

just that point

nowhere else

[alpine portal]

and -5

for the first function

[fringe stream]

and you should probably try -25/6

[alpine portal]

I don't get it... if f(x) has the constraint that it cant equal -5

[fringe stream]

not a mixed fraction

[alpine portal]

ill try

[solemn tiger]

Out of practice on composites, made a mistake.

The domain of the composite is still dependent on the original.

[alpine portal]

wtf

that worked

hahaha

[fringe stream]

yeah the thing probably just didnt recognize 4 1/6

[alpine portal]

-25/6 worked

[fringe stream]

because it looks like a 41 maybe?

just stick to fractions for online questions

[alpine portal]

yeah ahha fuck computer language

[solemn tiger]

I find that online hw systems tend to dislike mixed fractions

Although I think improper fractions are easier to do calculations with

[alpine portal]

But yeah the first constraint does carry over then

I thought this was a trick question

[fringe stream]

mhm

f(f(-5)) wont be defined bc f(-5) isnt defined itself

we learn something new everyday

[solemn tiger]

But only if the resultant function is a composite. If it's just a regular function you find on a problem, f(-5) is defined.

[fringe stream]

?

f(-5) for f(x) = x/(x + 5) is not defined?

[solemn tiger]

If we see f(f(x)) in some problem somewhere, f(-5) is defined

[alpine portal]

At the top it says composed

[solemn tiger]

If f(f(x)) weren't a composite*

[fringe stream]

oh well

yeah, ofc

[alpine portal]

thanks for the help guys anyways

[viscid thistle]

How does one factor nested trinomials with nested binomials as terms

[lost cipher]

factor out 3 and then expand?

if the middle term was (x+12) it would be easier