precalculus

[brisk forge]

im getting so lost rn lmao

sorry

[uncut mulch]

is the concept of the h confusing you?

what if it was any other variable

a^2/a

b^2/b

[viscid thistle]

Try replacing h with a number.

Would 2^2/2 = 1?

[brisk forge]

a'

[viscid thistle]

Or 3^2/3 = 1?

[brisk forge]

oo

ok

so it would just be h?

[uncut mulch]

yes

so going back, what's
3xh^2/h

[brisk forge]

3xh

[uncut mulch]

and what is
$\frac{h^3+3x^2h+3xh^2}{h}$

[obsidian monolith]

ramonov:

[brisk forge]

h+3x^2+3xh?

3x^2+3xh+h

[uncut mulch]

$\frac{h^3}{h} \neq h$

[obsidian monolith]

ramonov:

[brisk forge]

i dont get why not

if i separated it

like h/h, h/h, h/h

wouldnt it just be h

or 3h

[uncut mulch]

the ^3 only applies to the h on the top

$\frac{hhh}{h}$

[obsidian monolith]

ramonov:

[uncut mulch]

otherwise it would've been (h/h)^3 which is 1

[brisk forge]

ok so if i divide each of the numerators by h

itd just be h 3 times

3h right?

[uncut mulch]

no

[brisk forge]

wtf lmao

i do not get the concept

sorry. also ive been up for a while. this should be getting through to me

[uncut mulch]

2 ^ 3 = 2 * 2 * 2

[brisk forge]

yeah

[uncut mulch]

what happens when you divide that by 2

[brisk forge]

u get 1

[uncut mulch]

222 = 8

8/2 = 1???

[brisk forge]

oh i didnt know u solve it first

ok

so in the case of h

[uncut mulch]

you're not dividing every term

you divide the whole thing once

[brisk forge]

so if i divide h^3 by h how would that work

i cant solve that

hhh/h

= 3?

[uncut mulch]

no

cancel out an h from the top and bottom

or use index laws.
h^3/h = h^(3-1)

[brisk forge]

whaaat. i did that earlier and someone told me it was wrong

[uncut mulch]

what do you mean

[brisk forge]

https://media.discordapp.net/attachments/578040475928494092/619363537801838603/unknown.png

this was my first answer

[uncut mulch]

oh

that's the correct division
but you didn't take the limit h->0

[brisk forge]

ohhh how does that part work

[uncut mulch]

since h != 0, you were able to divide by h.

now what does that approach when h approaches 0

[brisk forge]

what is h!?

[uncut mulch]

a variable

[brisk forge]

no the h !=0

soz

[uncut mulch]

!= means doesn't equal to

[brisk forge]

oh ok

ok so if h is 0 its 3x^2+3x right

[uncut mulch]

3x*0 = ?

[brisk forge]

0

oh wtf

WOW

LOL

so its just 3x^2

[uncut mulch]

yes

[brisk forge]

jesus

THANK YOU LOL

i cant say thanks enough

i learned so much from all this

[viscid thistle]

Imagine knowing lims but not knowing this.

Shawty interesting fella.

[brisk forge]

LOL

[uncut mulch]

when someone said that that was wrong earlier,
why did you just abandon everything you knew about indices/division

[brisk forge]

idk honestly

ive been up for a while

like 27 hours or something

[uncut mulch]

goto sleep

[brisk forge]

gotta finish this hw lollll

[willow bear]

what have you tried & where are you stuck

[languid dust]

b

idk why they arent -2,1

[willow bear]

why what isn't "-2, 1"?

what are you asked for?

[languid dust]

remaining factor

[willow bear]

so you're saying you don't know why the remaining factor isn't "-2,1"?

am i interpreting what you're saying here correctly?

@languid dust

[languid dust]

what

does that even mean

yes

sure

i guess

[willow bear]

what does what mean

i was under the impression that those 2 messages i wrote were in plain english and so should not have posed a challenge to understand

@languid dust, can you write out your polynomial in factored form, using some sort of placeholder (like g(x)) for the unknown factor which you are tasked with finding?

hello?

[languid dust]

what

[willow bear]

what are you whatting at

are you unable to understand messages longer than a dozen words

[rare zephyr]

so someone asked me to solve this logarithm question that she can only remember but not refer it to

she stated that the question demands no calculators be involved in this question

[willow bear]

what is the question

[rare zephyr]

wait

im like writing this out

The first term is easy to solve

But what about the second one

is there a way to show my work for this

[uncut mulch]

how would you write 1/7 in index form?

[rare zephyr]

7^-1 ?

[uncut mulch]

where did the log_3 1 come from?

its more simplified when powers are brought down
you've also made things more complicated by evaluating 7^4

[rare zephyr]

Even if powered down

What should I do afterwards

[uncut mulch]

stop bringing the power up. you can leave the 4 as a coefficient

[rare zephyr]

log_3 7^4 is in decimals

[uncut mulch]

you didn't answer my previous question of where the log_3 1 came from

[rare zephyr]

log_3 (1/7)^4
= log_3 (1/7^4)
= log_3 1 - log_3 7^4

it IS possible to solve this without a calculator right....?

[uncut mulch]

i mean you could use the quotient rule like that but using the power rule,
log_3 (7^-1) = - log_3 (7)

log_3 1 is just 0

you didn't fix have the correct sign earlier

[rare zephyr]

oh yeah

got it

[uncut mulch]

don't bring the power up

and since log_3(7) can't be simplified, you just leave it like that

what would your final answer be

[rare zephyr]

With a calculator it should be 7.92

Yeah I'm stuck

[uncut mulch]

no

[rare zephyr]

what did i do wrong

[uncut mulch]

its more simplified when powers are brought down
stop bringing the power up.
don't bring the power up

also its sum to product

but you shouldn't be using that

since the term on the left was already simplified as 15

[rare zephyr]

I did thought of just leaving 15 be

But I'm actually stuck at answering that question for 3 days

Can you solve it @uncut mulch ?

[uncut mulch]

for simplifying,
you already had it at
15 + 4log_3(7)

[rare zephyr]

It's not solvable without a calculator, is it...

[uncut mulch]

you don't need to have it as a decimal for it to be solved

or unless you were supposed to get a nice answer, the question may have been transcribed incorrectly

[rare zephyr]

Well the girl that asked me to solve this didn't say what the answer was

[frozen needle]

"solve" is a very vague word

[rare zephyr]

she only got that question from memory

[frozen needle]

"solve: 23" doesn't make much sense as you can see

[uncut mulch]

if the answer was supposed to be 'nice',
the base in the 2nd log may have been 7

evaluate would be more appropriate here

[rare zephyr]

true

[viscid thistle]

why does spivak ask you to solve this if it only has complex roots

[frozen needle]

why not

[viscid thistle]

?

What complex roots.

There is an answer.

Hint: Complete the square.

[frozen needle]

any real x satisfies it, that's a good enough answer

[viscid thistle]

oh

Also if you're not sure if there's an answer use $sqrt\b^2-4ac$ to check if any function is factorable.

im dumb

i thought it was equal to x

However you do sqrt in that.

XD

[willow bear]

$\sqrt{b^2 - 4ac}$

[obsidian monolith]

Ann:

[viscid thistle]

Thanks <3

[crude hemlock]

Would the - and the + indicate “approaching from n direction”?

Or would it indicate “going in n direction”

[stuck lark]

yes, - means approach 1 from the left, + means approach 1 from the right

[crude hemlock]

Thanks

[stuck lark]

np

[viscid thistle]

im sorry if this sounds extremely stupid but how am i supposed to find x using only

this ?

[lucid crow]

@viscid thistle can you express 8 as two to the power of something something?

[viscid thistle]

i know

2^3

and then what

[lucid crow]

2^x < 2^3

the bases are the same

x < 3

and done

[viscid thistle]

is that rigorous enough ?

[primal karma]

it absolutely is

[viscid thistle]

alright

thank you both

[lucid crow]

well

if the bases are the same, then the only way the two expressions could be equal is for the powers also to be the same. That is:

x < 3

[viscid thistle]

that only works for n>1 right

bases n*

[primal karma]

for all i know it works for all positive real bases except 1, it's just a little bit diffrent

[viscid thistle]

0.99^googleplex is smaller than 0.98^2

so it has to be bigger than one

[primal karma]

i'ma give a little example just a second

you do need to change it a bit

but just a tiny bit

for example $(\frac{1}{2})^x<\frac{1}{4}$ you could write as $2^{-x} < 2^{-2}$ so therefore $-x<-2$ or by that $x>2$

[obsidian monolith]

Ꞡꭍî╥(│-│ḛƉ քîҳӘꭍş:

[primal karma]

works just as well, it only gets tricky with negative bases, 0 and 1

[viscid thistle]

that made things clearer

thanks for the help!

[primal karma]

No problem c:

[tidal rain]

Hey do i need to apply any properties here for question 7

[river coral]

How do I convert this expression into radical form? --> -4x^2/5y^3/5

[frozen needle]

it's hard to read if you don't use parentheses

[river coral]

-4x^(2/5)y^(3/5)

[frozen needle]

Ah, that's better

so what's giving trouble?

[river coral]

i tried doing this as my answer but it's wrong--> -4 root(2)(x^5) root(3)(y^5)

i just realized my mistake

its not root(2)(x^5) but it's root(5)(x^2)?

[frozen needle]

Yeah

[river coral]

so would the answer be the same as my wrong answer but to flip the 2 and the 5 (also the 3 and the 5)?

[frozen needle]

Yeah

[river coral]

ok thanks

[frozen needle]

🍻

[river coral]

help

How do you change this into exponential form? --> root(12)(a)(b^5)

[river coral]

nevermind i got it

[vale goblet]

i need help with this problem

[uncut mulch]

what have you tried?

[vale goblet]

34174.3

$34174.3

[uncut mulch]

how are you using the formula?

1 sec

[vale goblet]

so P = 8500

because it's the initial value

and since it's semi-annual, n = 6

r = .031

nt = 270

[uncut mulch]

semi annual means it applies twice

[vale goblet]

it should work

OHHHH

i didn't think about it hard enough

so it'd be 2(.031/6)

?

or just 12 on the bottom

[uncut mulch]

use n=2

[vale goblet]

oh

ok

i'll try that

uhhh

i got 540753.5955

[uncut mulch]

how are you entering in the values now?

[vale goblet]

8500(1 + 0.031/2)^270

wait

wait

wait

i forgot

i was still multiplying 45 by 6 on the top

33932.23159

yeah, that was right

also this one

but afk

[viscid thistle]

f(x) = -9e^(x-1)

f(-3)=-9e^(-3-1)

[vale goblet]

yeah so i assumed it'd be -9*10^-4

but it didn't like that

[viscid thistle]

no e is a number

[vale goblet]

it is

[viscid thistle]

,w e to 10 digits

[obsidian monolith]

Display results online and refine query

Requested by Babowwz#1394

[vale goblet]

but how does that work

is e always a predefined number

[viscid thistle]

f(-3)= -9/(e^4)

e is defined by a higher math process

[vale goblet]

bruh, we haven't learned this yet

[viscid thistle]

its ok

find e on calculator

-9/e^4

put it in

ez pz

[vale goblet]

but

[viscid thistle]

,w -9/(e^4)

[obsidian monolith]

Display results online and refine query

Requested by Babowwz#1394

[vale goblet]

OH WAIT

i forgot we have to uh

divide

because it's a negative exponent

how am i in pre-cal

[uncut mulch]

e^(-4) directly in the calc works too

[vale goblet]

ok yeah, i got the correct answer

thanks guys

[vale goblet]

how would i even start this

nvm i got it

[river coral]

Anyone knows how to do distributive law on this guy--> (x^(2/5)+2)^2

[stuck lark]

what did you try?

[river coral]

my answer came out like this: x^(4/5)+4^(2/5)+4 but its wrong

[stuck lark]

hmm close but not quite

[river coral]

so how would you solve it

[stuck lark]

do you remember "FOIL"?

[river coral]

yes

[stuck lark]

$(x^{2/5}+2)^2$

[obsidian monolith]

RokettoJanpu:

[stuck lark]

so it's the same as this

$(x^{2/5}+2)(x^{2/5}+2)$

[obsidian monolith]

RokettoJanpu:

[stuck lark]

what does the F in FOIL stand for?

[river coral]

uhhhhhhh

i forgot

[stuck lark]

"first" terms

[river coral]

oh ok

so you combine the two x^2/5?

[stuck lark]

yes multiply those

[river coral]

so it becomes x^4/5?

or am i adding it wrong

[stuck lark]

all good. now what does O stand for?

[river coral]

the outside?

[stuck lark]

yep. outer terms

[river coral]

combine the x^(2/5) with 2?

[stuck lark]

yes

[river coral]

so it becomes 2x^(2/5)?

[stuck lark]

ya

what does I stand for?

[river coral]

the inside?

[stuck lark]

ya. inner terms

[river coral]

combine another 2 and x^2/5?

[stuck lark]

ya

[river coral]

and its again 2x^(2/5)

[stuck lark]

k, what does L stand for?

[river coral]

idk

last?

[stuck lark]

ya

[river coral]

multiply the 2s?

[stuck lark]

ya

[river coral]

so its 4

[stuck lark]

now add up all the stuff you got from FOILing, what do you get?

[river coral]

so i have x^(4/5)+2x(2/5)+2x^(2/5)+4

so the middle what does it becomes 4^(2/5) or something else?

[stuck lark]

the terms in the middle, you're adding 2x^(2/5) + 2x^(2/5)

do you remember how that adds up?

[river coral]

so i add the whole number and add the x^(2/5)+(2/5)?

[stuck lark]

hmm lemme give you an example

does this easily add up?

$3x^2 + 7x^3$

[obsidian monolith]

RokettoJanpu:

[stuck lark]

as in, can you express this as only one term?

[river coral]

no because of the whole numbers

so you cant do anything

[stuck lark]

(actually it's because the exponents on the x's aren't the same)

[river coral]

oh

[stuck lark]

$3x^2 + 7x^2$

[obsidian monolith]

RokettoJanpu:

[stuck lark]

THIS you can add

[river coral]

i see

[stuck lark]

$= 10x^2$

[obsidian monolith]

RokettoJanpu:

[stuck lark]

see that?

[river coral]

yeah

[stuck lark]

same thing with 2x^(2/5) + 2x^(2/5), can you add that?

[river coral]

maybe?

so what do you do?

[stuck lark]

just like in the example i put above

[river coral]

so i can combine them?

[stuck lark]

$3x^2 + 7x^2 = 10x^2$

[obsidian monolith]

RokettoJanpu:

[stuck lark]

see how you can add these because the exponents of x are the same?

[river coral]

yeah

OH

[stuck lark]

so this is what you're adding

[river coral]

so its 4^(2/5)?

[stuck lark]

$2x^{2/5} + 2x^{2/5}$

[obsidian monolith]

RokettoJanpu:

[stuck lark]

not 4^(2/5)

[river coral]

is it just 2^(2/5)?

idk what im doing

well i know that if the base and the exponent is the same (the x^2) the whole number changes

cause if its not 4x^(2/5) which looks right, i dont know what makes it wrong

[stuck lark]

now THAT is right

[river coral]

wait letme look back wh-

OMFG i just forgot the fucking x

thanks

[stuck lark]

no problem man 🙂

[rose locust]

can anyone help me with trigonometric ratios?

[blazing parrot]

[river coral]

can you factor this equation any further?--> b^(3/2)+b

[proud sparrow]

sort of

[river coral]

and the answer says its b(___) (i have to put the answer inside the parentheses)

[proud sparrow]

So, you want to take a factor of b out

how do you do that?

[river coral]

the least common multiples?

so which is the smallest one the 3/2 or the 1/1?

[uncut mulch]

it already tells you tell factor is b, so you'll need to find
b^(3/2)/b and b/b

and that would be the hcf or gcd, not the lcm

[river coral]

oh

the directions wants the answer to be in positive exponents

[rose locust]

Can someone help with 6? thank you in advance!

[proud sparrow]

okay you have the quadrant

Check the reference angle

[rose locust]

Should be 60 degrees right?

which makes the angle in quadrant 3 = 240 degrees

and 300 in quadrant 4

[proud sparrow]

nope, not 60 degrees

well

you didn't find the correct angle

no

[uncut mulch]

it isn't a special triangle

[proud sparrow]

sin(x) is not equal to cos(x)/tan(x)

[rose locust]

I think I fixed the triangles but I’m not sure if I’m using the right one or if the angles are right. I think I’m supposed to use a 30/60/90 triangle so I filled in the angles as shown but I don’t think I’m on the right track

[uncut mulch]

it isn't a special triangle

how would you find an angle given a random ratio?
like sin(y) = 2.333/5

the question also tells you to round to 2.dp. so you angle shouldn't be a whole number

[rose locust]

well I would use sin-1 X -1/3 to find the angle, right? doing math at 2 am isnt the best time for me so I'm sorry if I'm slow at this

[proud sparrow]

???

arcsin -1/3?

[uncut mulch]

fix that notation
sin-1 X -1/3 t doesnt make any sense

[swift wagon]

arcsin(x)arccos(x) = -1

Is there any way to do this analytically?

[frozen needle]

arcsin(0)arccos(0) = 0

[swift wagon]

Yea

Wait I may of got it

[native sequoia]

solve for x, right

[frozen needle]

oh it wasn't a claim, I realized xD

[swift wagon]

Yep should I just use arccos(x) = Pi/2 - arcsin(x)?

[frozen needle]

yeah

[swift wagon]

Ok

[full garden]

guys i need help

whats the range for this functions

[proud sparrow]

@full garden What sort of values that this function take?

Analyze parts of it that you understand

like maybe the x-3 part

[full garden]

for the domain there's an asymptote on 3

[proud sparrow]

but we need the range

[full garden]

and the range could've been y does not equal to zero

but I dont know how the +5 would change the range

[proud sparrow]

okay

then analyse without the +5 first

can you do that?

[full garden]

ya the y cannot be zero

[proud sparrow]

So, in set notation, the range is {r real | r = y+5 where y is not 0}

do you agree?

[full garden]

ya

i mean i dont know what like this +5 does

is it like a transformation

[proud sparrow]

Is 1 in {r real | r = y+5 where y is not 0}?

[full garden]

will it be an asymtote

idkkk

[proud sparrow]

We need it to satisfy the condition to be in the set

[full garden]

do i solve for y?

[proud sparrow]

So, is 1 in the set?

[full garden]

ya

[proud sparrow]

You solve for which numbers are in the set

And why is it in the set?

yeah, you solve for y and find it is not 0

So, what sort of number would force y to be 0?

[full garden]

do i set x = 0

3

+3

[proud sparrow]

Let's check

{r real | r = y+5 where y is not 0}
3 = -2+5
so, here 3 is in the set

[full garden]

im sorry but how did you do that

[proud sparrow]

Okay, to check if a number is in the set, you need to solve for y and check it is not 0

So, what numbers are in the set?

[full garden]

all nubmers except for y

for 5 sorry

i think i got it

the horizontal asysmtote for the original equation without the +5 was 0

so 0 +5 = 5 so the new asysmtote is 5

so range y doesn't equal 5 ?

[proud sparrow]

yeah

it's all real numbers except 5

[full garden]

aight thank you

[tidal rain]

how to solve this??

[rigid beacon]

what have you tried?

[tidal rain]

isolating 1/r3

[hexed ermine]

Right

Now how do you turn 1/R3 to just R3

[hexed ermine]

Looks good

[tidal rain]

Thanks i forgot common denominator being a thing

and also i can only flip both sides of an equation if there is only one fraction on both sides?

[stuck lark]

You can always flip both sides if you really want to

[blazing parrot]

you can flip whenever but the entire thing has to be flipped

[tidal rain]

ah okk

[obsidian monolith]

RokettoJanpu:

$\frac{2}{a} = \frac{3}{b} + \frac{4}{c} \\\\ \frac{a}{2} = \frac{1}{\frac{3}{b} + \frac{4}{c}}$

[steel tulip]

How do you draw a tangent line

I know what it is but not how to draw it

[undone pawn]

you know how to find the tangent line equation?

the tangent is always like parallel to the direction of the curve at each point

[steel tulip]

Ok nevermind I'm stupid

You dont need to draw the tangent line I think

[undone pawn]

oh ok

wait what

you guys use commas for decimals

[steel tulip]

Haha

[vale goblet]

what

[heady jewel]

this isnt precalulculus

you can go to #prealg-and-algebra

[short sorrel]

exponential growth was covered in my precalc class.

either way, the borders are fuzzy, and saying "go to another channel" when they're so close is unhelpful and useless

[lean quest]

Hello, this is from UW precalc book and number 2 and 4 are really confusing me

nvm not number 4

[short sorrel]

for number 2, recall that v = d/t

now, set up the system:

$\frac{160}{60} \cdot v = d \ \ \frac{154}{60} \cdot (v+1) = d$

[obsidian monolith]

Namington:

[short sorrel]

do you see where i got this from?

the (160/60) is the time; 2 hours (120 minutes ) + 40 minutes, divided by 60 since we want hours, not minutes

(since our unit is km/hr)

the 154 comes from 160 - 6, since we took off 6 minutes

can you solve this system?

if your class uses decimals instead of fractions, feel free to rewrite the time as 2.6667 or whatever

but fractions are better

[lean quest]

Thanks for your response! I've come this far, but I get stuck at the solving part. If we set the two equations equal to each other, you cant isolate v

Oh, my bad, all I had to do was expand!

thanks for explaining things to me, it all makes sense now!

[pseudo sonnet]

anyone can check my work?

[trail badge]

@pseudo sonnet that's right yeah

[pseudo sonnet]

can you tell me how (x-10)^8 = 0

x = 10

dont you treat the 8 as if its just a 2 squared

but 10 isnt a perfect square

[trail badge]

(10-10)^8 = 0^8 = 0

[short sorrel]

$(10-10)^8 = 0^8 = 0$

[trail badge]

._.

Yeah...

[obsidian monolith]

Namington:

[trail badge]

Forgot $

[short sorrel]

dollar signs before and after

but yeah, basically

[trail badge]

Thanks

[short sorrel]

$(x-10)^8$ can be rewritten as $(x-10)(x-10)(x-10) \dots$

[obsidian monolith]

Namington:

[short sorrel]

8 times

each of those factors is 0 when x = 10

so the unique solution is x = 10

(expanding it like i did isnt necessary, but it might help you "visualize" why this is true)

[pseudo sonnet]

oh

im dumb lol

makes sense ty

[short sorrel]

\implies

[obsidian monolith]

Digicat195:

[pseudo sonnet]

ty

[trail badge]

Uw

[pseudo sonnet]

can anyone check this too

the problem is at the top

solution at bottom

work in middle

[viscid thistle]

Ye it's right.

You can check it yourself by plugging in your solutions.

[drowsy scaffold]

^

[pseudo sonnet]

how do i check

[viscid thistle]

So your result was $(1,\frac{3}{2}]$

[obsidian monolith]

leviosa:

[viscid thistle]

Since what you got was that 1 is not your solution.

You can plug it back into your parent function.

And you realize that it is not true.

Then you can check 3/2 and see if that works.

Yep, it equals to 1.

Then you can use 2, and no 1/2 is not greater than 1.

Now you know your answer is correct because your answer clearly tells you which numbers are right and which are wrong.

[viscid thistle]

HI
I'm having trouble with this question
| sqrt(5) - 5 |
It says to rewrite the expression without using the absolute ­value symbol

5-sqrt(5)

;-;

[native timber]

evaluate $\sqrt{5} - 5$ but remove the negative sign after

[obsidian monolith]

Jiramide:

[viscid thistle]

Or just do $5-\sqrt{5}$

[obsidian monolith]

leviosa:

[native timber]

yeah

[viscid thistle]

is that correct?

i thought it was an inequality 😿

You only gave an expression.

So if it's an inequality we wouldn't know.

[native timber]

$\mid a\mid = \begin{cases}a \le 0 &-a\otherwise &a\end{cases}$

[obsidian monolith]

Jiramide:

[viscid thistle]

so the correct answer would be 5 - sqrt(5)

Yes.

okay

how about | 5 -23 |

its asking for the same thing

so 23-5?

Yes.

Just try it yourself.

Is $|5-23|=23-5$

[obsidian monolith]

leviosa:

[viscid thistle]

You can plug it into a calculator.

[native timber]

given $\mid a - b \mid$, if $b \ge a$ then its equal to $b - a$

[obsidian monolith]

Jiramide:

[viscid thistle]

^

Precisely.

<3

How about...

| x-2 | if x < 2

[short sorrel]

if x < 2, is x - 2 positive or negative?

[viscid thistle]

negative

The think you just stated looks something like this: $|-|x|-2|$

Which will always be negative.

And since x is always negative, what would you do to x to make it always positive.

Uh no I didn't write that

I wrote |x-2|

I'll be back in 20 minutes

[short sorrel]

@viscid thistle I'm... not really sure what you're doing

you cant just make the x posiitve

you have to make the whole expression positive

thats how absolute values work

[viscid thistle]

I'm just doing it mentally lmao.

[obsidian monolith]

leviosa:

[viscid thistle]

I meant for him to do it something like this: $-2-x$

[obsidian monolith]

leviosa:

[viscid thistle]

Since x is always negative.

In the end it'll be positive.

Worded it badly, mb.

um sorry I still dont understand it

[short sorrel]

i... don't understand either

i think you're changing the function, @viscid thistle

[viscid thistle]

'

Using this.

[short sorrel]

right

but you didnt write b - a

you wrote -a - b

[viscid thistle]

.>

Because.

x is always negative.

[short sorrel]

?????

[viscid thistle]

Sorry I wrote it immaturely.

Wait.

[short sorrel]

you wrote a different function

[viscid thistle]

Shit.

| x-2 | if x < 2

Okay so we have that.

So if you want to make it positive.

Without the use of abs value.

Just do 2-x (x<2)

That's what I meant.

[short sorrel]

there we go

[viscid thistle]

.>

[short sorrel]

basically, what $|stuff|$ does is "looks at" that stuff

[obsidian monolith]

Namington:

[viscid thistle]

[short sorrel]

and says "if this is negative, make it positive"

"otherwise dont change it"

the way it "makes it positive" is by multiplying by -1

now, we know that x - 2 is ALWAYS negative when x < 2

so |x - 2| will basically multiply -1 in

to write this more explicitly, we just write

-1 * (x-2)

which is the same as -x + 2

or, if you prefer, 2 - x.

[viscid thistle]

Oh.

Wow I'm dumb.

[fresh kelp]

https://gyazo.com/1b13c2f0402c5fce2f7a76ff2768550e

i know the numerator cancels to 1

how do u distribute the denominator? or simplify it

[viscid thistle]

What's the common term in the numerator and the denominator?

[uncut mulch]

the numerator doesn't cancel to 1

[fresh kelp]

errr

uhhh

h

[viscid thistle]

Yes.

Just h.

[uncut mulch]

does that resolve your issue?

[fresh kelp]

so

we first conjugate right

heres the full example problem im on

https://gyazo.com/c92588ac56b6cfea5aadb77d99af0221

[viscid thistle]

Yes.

[uncut mulch]

yeh. simplifying, you'd have h on the numer and denom

which means you can :

[fresh kelp]

so h is on top and bottom

im sorry is there a word for this type of problem? i will google a video, im not understanding it quite well

[uncut mulch]

difference quotient

you reached h/((h (sqrt(h+16)+4))
what would be the next step?

[fresh kelp]

oh the problem is where does the top h come from

[uncut mulch]

difference of two sqrs. Expand the top

sqrt(16+h)^2 = 16 + h

[fresh kelp]

mind if i draw it

[viscid thistle]

Draws an equation.

[fresh kelp]

D:

write it out

[uncut mulch]

yeh

[viscid thistle]

Playing a game: Adobe Photoshop CC 2018
Draws an equation.
Uses Photoshop to write equation.

[fresh kelp]

;-;

stop bully

https://gyazo.com/41df6eb732f723f4dcc90049608958f7

this is conjugating right? D:

[viscid thistle]

I believe so.

Yes.

It is.

[fresh kelp]

oki so i have to add an h on the right side?

[uncut mulch]

nope

[fresh kelp]

;-;

[viscid thistle]

Add h where.

[uncut mulch]

expand the numerator

[fresh kelp]

like a common denominator? ;-;

[uncut mulch]

as in
sqrt(16+h)^2 - sqrt(16)^2

[fresh kelp]

no like can i multiply straight accross? ;-;

[uncut mulch]

from diff of 2 squares

well that's what you're doing by multiplying top/bottom by the conjugate

[fresh kelp]

;-;

like this

thats how they got it in the example

right

OH

OH

I GOT IT

expand thew numerator

that was the issue

omg

[main vessel]

A bit low effort but is 3 correct? Wanted to make sure

[proud sparrow]

roots are correct

What's the vertex?

V=(13/3, 64/3)?

yeah, that looks correct

y intercept is -35

yeah

[main vessel]

Great thanks

[viscid thistle]

How can i find a function that satisfies f(x(x+1)/2)=x^2?

[proud sparrow]

@viscid thistle Many functions satisfy that

if x is positive

but if x can be positive or negative, then no functions satisfy that

[viscid thistle]

x is supposed to be a natural number

[proud sparrow]

Oh oka

So, just by plugging in x, what do you get?

Try a couple of examples

[viscid thistle]

f(1)=1, f(3)=4, f(6)=9, f(10)=16,...

[proud sparrow]

So, can you guess the function from this?

Another way you can try is calculating x from x(x+1)/2

[viscid thistle]

x^2+x-2s=0?

s=x(x+1)/2 and i solved for x is that what you mean

it's almost linear

https://www.desmos.com/calculator/gmdvtitrem

Is there like a general expression for all the functions that satisfy that relation?

x,x+1,x+3,x+6,x+10,

[proud sparrow]

there would be

I guess

Maybe you can try calculating x from x(x+1)/2, since from there you can easily calculate x^2

[viscid thistle]

i didn't get that

wdym

[proud sparrow]

Given that x(x+1)/2=120, can you calculate x?

[viscid thistle]

yeah x=15

[proud sparrow]

but how do you do that in general, if not guessing?

[viscid thistle]

solve the quadratic

wait

s^2=x(x+1)/2

then solve for x

is this what you mean

that's not it

@proud sparrow

[proud sparrow]

no

given that x(x+1)/2=A, solve for x

[fallen cloud]

That's just quadratic formula

[viscid thistle]

x^2+x-2s=0?

[fallen cloud]

Yeah

[viscid thistle]

so the function is the positive solution to that

(sqrt(8s+1)-1)/2

?

f(x)=(sqrt(8x+1)-1)/2

wait that's not right

[fallen cloud]

8x+1 in the sqrt

[viscid thistle]

yeah sorry hold on

so all of that squared

ohhhhhhhhhh

[fallen cloud]

8s+1 in the

Sqrt

[viscid thistle]

yeah but the function is f(x) so i wrote it in terms of x

so in conclusion the function is

f(x) = ((sqrt(8x+1)-1)^2)/4

where x is a natural number

cool and wolfram agrees

thank you both

is there another function that satisfies the relation

?

[proud sparrow]

well, since f(x(x+1)/2)=x^2 only matters if x is an integer...

think about what happens when x is not an integer

you can push the function somewhere else

[viscid thistle]

So just a positive number

I don't understand why that would make such a difference

[proud sparrow]

Well, if x is a positive integer ONLY

then f(2) is not specified

@viscid thistle

[willow bear]

can i have the exact statement of the problem?

[viscid thistle]

Well it's basically find all functions f(x) such that f(x(x+1)/2)=x^2 where x is a natural number

@proud sparrow oh i get it

[willow bear]

...i didn't ask for BASICALLY

i asked for the EXACT statement of the problem

do you actively choose to neglect the most important word in my message

[viscid thistle]

yeah it's not homework i'm just curious about smth i can't really get more specific than that

[viscid thistle]

Hi, how would I solve this inequality
(1/x) < 4

I got (1/4) < x

[royal gull]

[proud sparrow]

are you sure x is positive @viscid thistle

[viscid thistle]

it would be (-infinity, 1/4)

[proud sparrow]

NOPE

can it be 1/5?

[viscid thistle]

oh my bad

(1/4 , infinity)

sorry, just doing this in my head

[proud sparrow]

can it be -1?

[viscid thistle]

nope

[proud sparrow]

are you sure?

is (1/(-1)) < 4?

[viscid thistle]

yes

-1 is less than 4

[proud sparrow]

so?

[viscid thistle]

it would be negative infinity to 1/4

please tell me if im wrong...

[willow bear]

yes you are

[proud sparrow]

the solution set doesn't need to be a contiguous interval

[willow bear]

it's not a single interval

[viscid thistle]

okay. how would i approach it?

this question i mean
(1/x) < 4

[willow bear]

have you ever solved rational inequalities before?

[viscid thistle]

Yes, but I've forgotten how to do them

[serene heath]

your solution would work if x was always positive

but it doesnt have to be

[willow bear]

then maybe review rational inequalities & come back to this afterwards

[viscid thistle]

I know its switching places with the "4" and "x"

[willow bear]

no

it's not

[main vessel]

This is the kinda shit I see on SAT tests. I'm in Calc and I still never really learned how to do this one

[river kindle]

I'm assuming problem 14. Generally, you look for the lowest common denominator. Alternatively, multiply everything by x-6, then solve for x

[main vessel]

Everything, including the 1?

[river kindle]

everything, so yes.

even 0

[main vessel]

Gotcha, just making sure

[river kindle]

surprised you haven't learned it before. I figured knowing this would be necessary for calculus.

[main vessel]

err maybe I learned about it once but I dont remember learning about it extensively in pre calc

[river kindle]

Perhaps. My experience may be different as I took a break from education after high school, and going back, they forced me to take pre-alge, college algebra, and pre-calc before I can do anything else at my community college, so I feel like it's all been rigorously practiced for me. 😛

[main vessel]

Yeah I mean I never even learned about e and I have a few problems on my pre calc review sheet with it

[river kindle]

e? Natural log e?

[stuck lark]

the one and only E

[main vessel]

yeah I assume. I WAS informally taught about it as a trick to figuring out one or two AP Chem questions but that was it

Dont really know much about it

other than it's the opposite of ln or something

[main vessel]

I get this right? 🤔

10

[willow bear]

don't write your t's like this, it makes them too similar to +

but other than that yes everything is fine

[main vessel]

Alright thx, I'm not very good at ts haha

[main vessel]

This question makes me wanna hit it with a stick 😠

(14)

[river kindle]

6/x should be 6(-6).

$(6/x) * x-6$ x is divided out by the x in the denominator

[obsidian monolith]

neveza:

[river kindle]

Actually, nevermind, I think I wrong on that assumption

[main vessel]

Oh err I get what you mean but I thought you couldn't do that

[river kindle]

You may need to multiply everything by x(x-6) when I think about it

[main vessel]

OH shit I did learn that last year

Multiplying by both denominators at once

[river kindle]

Right.

[remote citrus]

Can someone help with d?

[viscid thistle]

So we have 20 degrees.

And you want to A.) Find the coterminal and B.) Convert it back to radians.

Both of these concepts you should know.

[remote citrus]

You

Are

A

Legend

[uncut mulch]

it's not 20deg. It should be 20radians.

[vague zephyr]

i just started college and this is a question on my homework, the first and last answers are correct, the second one is half correct, im not sure how to solve this

[viscid thistle]

@vague zephyr (-8, inf)

[vague zephyr]

i think i tried it but ill check again

no, its incorrect

[uncut mulch]

no

recheck your intervals

[vague zephyr]

i know that (-inf,-8) is correct, not sure about the other part

[uncut mulch]

how are you using orC

[vague zephyr]

im graphing this and i only see A and C overlapping so i thought the answer would be [-5,inf) but this is incorrect

[stuck lark]

does $\cup$ mean take what's overlapping? @vague zephyr

[obsidian monolith]

RokettoJanpu:

[vague zephyr]

no

[stuck lark]

so tell us what does it mean

[uncut mulch]

just that or did you include (-8,inf) in the answer?

[vague zephyr]

U means that its supposed combine all of the numbers within the sets

[stuck lark]

what's your answer then?

[vague zephyr]

i dont know, thats why i asked the question

[uncut mulch]

what do you currently think it is?

[vague zephyr]

well i know that (-inf,-8) is correct, but i do not know how to solve for the next set

[uncut mulch]

can you clarify what you meant by

i thought the answer would be [-5,inf) but this is incorrect

[vague zephyr]

i was only solving for AUC not AUBUC

[uncut mulch]

as in you only entered [-5, inf) into the answer box?

[vague zephyr]

no, (-inf,-8)U[-5,inf)

i've never had to solve intervals and sets before and vaguely have information from youtube, but this confuses me because i cant find where to solve for 3 union sets

[uncut mulch]

that should be correct

[vague zephyr]

but i submitted this and it wasnt

do you know what else it could be?

i took a pre cal class in high school and we didnt cover this

[river kindle]

Have you mapped it all on a number line? Might help

[vague zephyr]

i did

[tribal sun]

Hello

I need to know how to do 79-89

I don’t want to know the answer even if it says it right there I just want to learn

Pls anyone teach me 😦

[proud raven]

so

can you write $\frac{f}{g}(x)$ in terms of g's and x's?

[obsidian monolith]

jan Niku:

[proud raven]

you need to see that you have an equation

$\frac{f(x)}{g(x)}\cdot (x) = \frac{x+1}{x^2-x}$

[obsidian monolith]

jan Niku:

[proud raven]

and you know f(x) in terms of x

so write the entire thing in terms of g's and x's, and solve for g(x) @tribal sun

[tribal sun]

Ok lemme right it out

[proud raven]

is that for 79?

[tribal sun]

Yea I tried to do it last night

[proud raven]

there is no difference quotient in 78

[tribal sun]

79

[proud raven]

im talking about 78 atm

did you not need help with that problem

okay

oh

youre totally right and i misread 😄

[tribal sun]

Lmao

[proud raven]

gimme a sec

so were looking at 79?

[tribal sun]

Yes

[proud raven]

it doesnt look like you have 4x+3 on that paper

[tribal sun]

Oh shet

[proud raven]

fwiw i was helping some in real life with these most of the day yesterday and 90% of the time its just being disorganized and skipping steps that leads to problems

[tribal sun]

79 is the bottom one

[proud raven]

do you know how to do these

[tribal sun]

Yea isn’t it that formula

[proud raven]

yup 😄

try writing it out

[tribal sun]

I don’t know how to implement the formula

Lol

[patent beacon]

Above you had no problem identifying f(x + h), where are you stuck?

[tribal sun]

What becomes of the 4(x)

[proud raven]

youre trying to skip steps

the first step is write the difference quotient

i assume you are at least

you have f(x+h) and f(x) written there

[tribal sun]

Yes

[proud raven]

the first step is to write out the difference quotient

so just write out that formula with everything you have there

[tribal sun]

?

[proud raven]

you have the difference quotient written just there

on the right of the two formulas

[proud raven]

you also have f(x+h)

you'll know youre done when you have nothing but x's and h's as variables

[tribal sun]

Can you give me the steps I lose it after the substitution

[proud raven]

you need to write out what f(x) is from the problem

then write f(x+h)

then plug those in to $\frac{f(x+h)-f(x)}{h}$

[obsidian monolith]

jan Niku:

[proud raven]

dont do any simplification

[vague zephyr]

@uncut mulch the answer you helped me with was incorrect

[tribal sun]

I don’t get it

[patent beacon]

You've written out f(x + h), you've written out f(x). Now write out f(x + h) - f(x) / h.

Specifically, don't cancel or simplify anything - just write everything on one line

[uncut mulch]

unless i fked up, it should be

[vague zephyr]

it isnt right, the teacher said it wasnt

[proud raven]

You wrote this in an earlier answer @tribal sun , did you understand where this came from?

[tribal sun]

Because all x’s turn into (x+h)

???

[proud raven]

right

[uncut mulch]

did the teacher say why it wasn't?

[tribal sun]

I did not get tha part

[vague zephyr]

he did not, he just said the second portion was incorrect

[proud raven]

what do you mean

are you confused about why it equals that? Or why you need it

[gaunt mural]

How do you know the difference between a horizontal stretch and a vertical compression and vice versa

Whenever I watch a video or graph them they seem the same and I can't tell the difference

[proud raven]

@gaunt mural it's not always clear on every graph

try it on something like cos(x)

[gaunt mural]

This is kinda what I figured out

idk if it is right tho

[proud raven]

its been a while since algebra but iirc sometimes theres no difference